A helicopter lifts a 85 kg astronaut 12 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/12. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed? (a) Number ___________ Units _____________
(b) Number ___________ Units _____________
(c) Number ___________ Units _____________
(d) Number ___________ Units _____________

Answers

Answer 1

A helicopter lifts a 85 kg astronaut 12 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/12.(a)The work done on the astronaut by the force from the helicopter is 85 kg × 9.81 m/s² × 12 m=9930.6 J.(b)the work done on the astronaut by the gravitational force is = -9930.6J(c)Kinetic Energy = 9930.6J(d)v ≈ 15.26 m/s

(a) To calculate the work done on the astronaut by the force from the helicopter, we can use the formula:

Work = Force × Distance

The force from the helicopter can be calculated using Newton's second law:

Force = Mass × Acceleration

Given that the mass of the astronaut is 85 kg and the acceleration is g/12 (where g is the acceleration due to gravity, g = 9.81 m/s²), the force from the helicopter is:

Force = 85 kg × (g/12) m/s²

The displacement of the astronaut is given as 12 m.

Substituting the values into the work equation:

Work = (85 kg × (g/12) m/s²) × 12 m

Simplifying the equation, we have:

Work = 85 kg × g m/s² × 12 m

The units for work are Joules (J).

Therefore, the work done on the astronaut by the force from the helicopter is 85 kg × 9.81 m/s² × 12 m J.

(a) Number: 9930.6

Units: Joules (J)

(b) The work done by the gravitational force can be calculated in the same way. The force of gravity can be calculated as:

Force_gravity = Mass × Acceleration_due_to_gravity

Given that the mass of the astronaut is 85 kg and the acceleration due to gravity is 9.81 m/s², the force of gravity is:

Force_gravity = 85 kg × 9.81 m/s²

Since the displacement is vertical and the force of gravity is acting in the opposite direction to the displacement, the work done by gravity is:

Work_gravity = -Force_gravity × Distance

Substituting the values:

Work_gravity = -(85 kg × 9.81 m/s²) × 12 m

The units for work are Joules (J).

Therefore, the work done on the astronaut by the gravitational force is -(85 kg × 9.81 m/s² × 12 m) J.

(b) Number: -9930.6

Units: Joules (J)

Note: The negative sign indicates that work is done by the gravitational force in the opposite direction to the displacement.

(c) Just before she reaches the helicopter, her potential energy is converted into kinetic energy. Since the work done by the helicopter and the gravitational force cancel each other out, her total mechanical energy (potential energy + kinetic energy) remains constant. Therefore, her potential energy at the start is equal to her kinetic energy just before reaching the helicopter.

Potential Energy = m×g×h

Given that the mass of the astronaut is 85 kg, the acceleration due to gravity is 9.81 m/s², and the height is 12 m, her potential energy is:

Potential Energy = 85 kg × 9.81 m/s² × 12 m

The units for energy are Joules (J).

Therefore, The kinetic energy just before reaching the helicopter is also:

Kinetic Energy = 85 kg × 9.81 m/s² × 12 m J.

(c) Number: 9930.6

Units: Joules (J)

(d) To find her speed just before reaching the helicopter, we can equate her kinetic energy to the formula for kinetic energy:

Kinetic Energy = (1/2)mv²

where m is the mass and v is the speed.

Substituting the values:

9930.6 J = (1/2) × 85 kg × v²

Simplifying the equation:

v² = (2 × 9930.6 J) / (85 kg)

v² = 233.25 m²/s²

Taking the square root of both sides:

v ≈ 15.26 m/s

(d) Number: 15.26

Units: meters per second (m/s)

To learn more about Newton's second law visit: https://brainly.com/question/25545050

#SPJ11


Related Questions

Write the 4-momentum P = (5 , pc) of E a particle of mass m in terms of its V rapidity defined by ?

Answers

The 4-momentum of a particle E with mass m can be expressed as P = (5, pc) in terms of its rapidity V.

The 4-momentum of a particle is a four-component vector that describes its energy and momentum in the context of special relativity. It is denoted as P = (E, pc), where E is the energy of the particle and pc represents the momentum in the x, y, and z directions.

In terms of the rapidity V, which is defined as the hyperbolic tangent of the particle's velocity v, we can express the energy E as a function of the rapidity.

The relationship between rapidity and velocity is given by the equation,

V = tanh⁻¹(v), where v is the velocity of the particle.

Solving for v, we find v = tanh(V).

To obtain the 4-momentum in terms of rapidity, we first express the energy E in terms of the particle's rest mass m and its velocity v using the relativistic energy-momentum relationship:

E = γmc²,

where γ is the Lorentz factor γ = 1/√(1 - v²/c²).

Substituting v = tanh(V), we can rewrite γ as γ = cosh(V).

Finally, we obtain the 4-momentum as P = (E, pc) = (γmc², γmvc), where c is the speed of light.

Simplifying this expression, we have P = (5, mc sinh(V)c), where sinh(V) represents the hyperbolic sine of the rapidity V.

Therefore, the 4-momentum of the particle E in terms of its rapidity V is P = (5, pc) = (5, mc sinh(V)c), where mc represents the magnitude of the particle's momentum in the x, y, and z directions.

To learn more about Lorentz factor visit:

brainly.com/question/31962456

#SPJ11

Alternating current have voltages and currents through the circuit elements that vary as a function of time. In many instances, it is more useful to use rms values for AC circuits. Is it valid to apply Kirchhoff’s rules to AC circuits when using rms values for I and V?

Answers

Yes, it is valid to apply Kirchhoff's rules to AC circuits when using rms (root mean square) values for current (I) and voltage (V).  Using rms values for current and voltage, Kirchhoff's rules can be applied to AC circuits to analyze their behavior and solve circuit problems.

Kirchhoff's rules, namely Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL), are fundamental principles used to analyze electrical circuits. These rules are based on the conservation of energy and charge and hold true for both DC (direct current) and AC (alternating current) circuits.

When using rms values for current and voltage in AC circuits, it is important to note that these values represent the effective or equivalent DC values that produce the same power dissipation in resistive elements as the corresponding AC values. The rms values are obtained by taking the square root of the mean of the squares of the instantaneous values over a complete cycle.

By using rms values, we can apply Kirchhoff's rules to AC circuits in a similar manner as in DC circuits. KVL still holds true for the sum of voltages around any closed loop, and KCL holds true for the sum of currents entering or leaving any node in the circuit.

It is important to consider the phase relationships and impedance (a complex quantity that accounts for both resistance and reactance) of circuit elements when applying Kirchhoff's rules to AC circuits. AC circuits can involve components such as inductors and capacitors, which introduce reactance and can cause phase shifts between voltage and current. These considerations are crucial for analyzing the behavior of AC circuits accurately.

To know more about Kirchhoff's voltage law (KVL)

https://brainly.com/question/30400751

#SPJ11

A student gets her car stuck in a snow drift. Not at a loss, having studied physics, she attaches one end of a rope to the vehicle and the other end to the trunk of a nearby tree, allowing for a small amount of slack. The student then exerts a force F on the center of the rope in the direction perpendicular to the car-tree line as shown. Assume equilibrium conditions and that the rope is inextensible. How does the magnitude of the force exerted by the rope on the car compare to that of the force exerted by the rope on the tree? 1. ∣F t

∣=2∣F c

∣ 2. Cannot be determined 3. ∣F t

∣>∣F c

∣ 4. ∣F t

∣=∣F c

∣=T 5. ∣F t

∣<∣F c

∣ 004 (part 2 of 2) 10.0 points What is the magnitude of the force on the car if L=19.7 m,d=2.26 m and F=596 N ? Answer in units of N.

Answers

The magnitude of the force exerted by the rope on the car is equal to the force exerted by the rope on the tree. The correct option is 4

This is because the system is in equilibrium, meaning there is no net force acting in any direction. In equilibrium, the tension in the rope is the same throughout its length.

∣Ft∣ = ∣Fc∣ = T, where T represents the tension in the rope.

Given the values L = 19.7 m, d = 2.26 m, and F = 596 N, the magnitude of the force on the car (Fc) is equal to the tension in the rope (T), which is 596 N. Both the car and the tree experience the same magnitude of force due to the inextensible nature of the rope and the equilibrium conditions. Therefore, the correct option is 4.

To know more about equilibrium, here

brainly.com/question/30694482

#SPJ4

Three charged conducting metal balls are hanging from non-conducting strings. Initially, ball #1 has a charge of -12 uc, ball #2 has 22 uC, and ball #3 has -11 PC. Ball #1 is brought in contact with ball #2 and then the two are separated. Ball #2 is then moved over and brought into contact with ball #3, after which the two are separated. What are the final charges on each ball?

Answers

The final charges on each ball are as follows:

Ball #1: 10 μC

Ball #2: -1 μC

Ball #3: -1 μC

To determine the final charges on each ball, we need to consider the transfer of charge when the balls come in contact with each other. When two conductive objects come in contact, charge can flow between them until they reach equilibrium.

Let's analyze the situation step by step:

Step 1: Ball #1 (-12 μC) is brought in contact with Ball #2 (22 μC).

When the two balls touch, electrons will flow from the negatively charged Ball #1 to the positively charged Ball #2 to equalize the charge distribution.

The net charge after contact will be the sum of the initial charges on Ball #1 and Ball #2.

Net charge = -12 μC + 22 μC = 10 μC

Ball #1 and Ball #2 now have the same charge of 10 μC each.

Step 2: Ball #2 (10 μC) is moved over and brought into contact with Ball #3 (-11 μC).

When the two balls touch, charge will flow to equalize the charge distribution.

Since Ball #2 has a higher charge, electrons will flow from Ball #2 to Ball #3.

The net charge after contact will be the sum of the initial charges on Ball #2 and Ball #3.

Net charge = 10 μC - 11 μC = -1 μC

Ball #2 now has a charge of -1 μC, and Ball #3 has a charge of -1 μC.

Step 3: Ball #1 (10 μC) is separated from Ball #2 (-1 μC).

The charges remain unchanged since they are no longer in contact.

To knwo more about electrons

https://brainly.com/question/12001116

#SPJ11

After a bomb at rest explodes into two unequal fragments, the more massive fragment has the same kinetic energy as the less massive fragment. more kinetic energy than the less massive fragment. less kinetic energy than the less massive fragment.

Answers

When a bomb at rest explodes into two unequal fragments, the more massive fragment has less kinetic energy than the less massive fragment.

According to the law of conservation of momentum, the total momentum before and after the explosion must be the same. In this case, since the bomb is initially at rest, the total momentum before the explosion is zero. After the explosion, the two fragments move in opposite directions, but their combined momentum must still add up to zero.

Since momentum is the product of mass and velocity, if one fragment has a greater mass, it must have a lower velocity to maintain the total momentum at zero. As kinetic energy is proportional to the square of velocity, the more massive fragment will have a lower kinetic energy compared to the less massive fragment.

This phenomenon can be explained by the conservation of energy. The initial energy of the bomb is stored in the form of chemical potential energy. When the bomb explodes, this energy is converted into the kinetic energy of the fragments. However, due to the unequal masses, the less massive fragment receives a greater share of the initial energy, resulting in a higher kinetic energy.

Learn more about kinetic energy here:

https://brainly.com/question/999862

#SPJ11

The boiling point of helium at one atmosphere is 4.2 K.What is the volume occupied by the helium gass due to the evaporation of 10 g of liquid helium at 1 atm of pressure for the following temperatures a) 4.2 K b) 293 K A cubic metal box with sides of 20 cm contains air at a pressure of 1 atm and a temperature of 300 K. The box is sealed so that the volume is constant, and it is heated to a temperature of 400 K. Find the net force on each wall of the box.

Answers

2.5 mol of helium occupies a volume of 22.4 L × 2.5 = 56 L. The volume of the helium gas is approximately 61.3 L. The net force on each wall of the box is approximately 2355 N.

a) The boiling point of helium at one atmosphere is 4.2 K. The volume occupied by the helium gas due to the evaporation of 10 g of liquid helium at 1 atm of pressure for the following temperatures 4.2 K can be calculated as follows:

Mass of liquid helium, m = 10 g

Molar mass of helium, M = 4 g mol^(-1)

Number of moles, n = (10 g) / (4 g mol^(-1)) = 2.5 mol

Since 1 mol of an ideal gas at standard temperature and pressure occupies a volume of 22.4 L, therefore 2.5 mol of helium occupies a volume of 22.4 L × 2.5 = 56 L.

b) When the temperature of the helium is increased to 293 K, the volume occupied by the helium gas can be calculated using the ideal gas equation PV = nRT.

P = 1 atm

V = ?

n = 2.5 mol

R = 8.314 J mol^(-1) K^(-1)

T = 293 K

Therefore, V = (nRT) / P = (2.5 mol × 8.314 J mol^(-1) K^(-1) × 293 K) / (1 atm) ≈ 61.3 L

The volume of the helium gas is approximately 61.3 L. Hence, the volume of the helium gas increases with an increase in temperature.

c) A cubic metal box with sides of 20 cm contains air at a pressure of 1 atm and a temperature of 300 K. The box is sealed so that the volume is constant, and it is heated to a temperature of 400 K. The net force on each wall of the box can be calculated as follows:

Initial pressure, P1 = 1 atm

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Volume, V = (20 cm)^3 = (0.2 m)^3 = 0.008 m^3

The final pressure, P2, can be calculated using the ideal gas equation:

P1V1 / T1 = P2V2 / T2

P2 = P1V1T2 / V2T1

P2 = (1 atm × 0.008 m^3 × 400 K) / (0.008 m^3 × 300 K) ≈ 1.33 atm

The change in pressure, ΔP, can be calculated using the equation:

ΔP = P2 − P1

ΔP = 1.33 atm − 1 atm = 0.33 atm

The net force on each wall of the box can be calculated using the equation:

Fnet = PΔA

= ΔPΔA

= ΔP × (2lw + 2lh + 2wh)

where l, w, and h are the length, width, and height of the box, respectively. Since the box is cubic, l = w = h = 20 cm = 0.2 m, therefore,

Fnet = ΔP × (2lw + 2lh + 2wh)

= (0.33 atm × 101325 Pa/atm) × (2 × 0.2 m × 0.2 m + 2 × 0.2 m × 0.2 m + 2 × 0.2 m × 0.2 m)

≈ 2355 N

The net force on each wall of the box is approximately 2355 N.

Learn more about boiling point: https://brainly.com/question/40140

#SPJ11

A machinist bores a hole of diameter \( 1.34 \mathrm{~cm} \) in a Part \( A \) steel plate at a temperature of \( 27.0^{\circ} \mathrm{C} \). You may want to review (Page) What is the cross-sectional

Answers

The problem is a case of linear expansion of solids. If there is a change in temperature in an object, then the length of the object also changes. And in this situation, the diameter of the hole changes. The diameter of a hole is directly proportional to the length of the plate. Hence, the formula for this situation would be ΔL=αLΔT

Where, ΔL is the change in length of the plate, L is the initial length of the plate, ΔT is the change in temperature of the plate, and α is the coefficient of linear expansion of the plate.

The formula for the diameter of the hole would beΔd=2αLΔTwhere, Δd is the change in diameter of the plate.

It is given that the initial diameter of the hole, d = 1.34 cm, the initial temperature, T = 27 °C, ΔT = 80 °C

Therefore, the change in diameter is,Δd = 2αLΔTWe know that steel is a metal and its coefficient of linear expansion, α is 1.2 × 10^(-5) K^(-1).

The value of L is not given.

So, let's assume that the coefficient of linear expansion of the steel is constant and also the value of L is constant.

Δd = 2αLΔTΔd

= 2 × 1.2 × 10^(-5) × L × 80Δd

= 1.92 × 10^(-3) L

The value of L can be calculated as,

L = Δd / (1.92 × 10^(-3))L = 0.7 m = 70 cm

Therefore, the length of the steel plate is 70 cm.

Thus, the answer is: The length of the steel plate is 70 cm.

Learn more about change in temperature here

https://brainly.com/question/27988898

#SPJ11

The switch is closed for a long time. It opens at t-0. i) Find i, (0+) and v₂ (0+) [3 pts] X1=0 692 12 V 2H 0.4 F For t > 0, what kind of system response does the series RLC circuit produce for i(t)? (Underdamped, overdamped, critically damped). Also, express the form of the solution. Find di(0*) and dv (0*) dt dt Iz(t) 492 :ve(t)

Answers

The current in the series RLC circuit is given by the equation i(t) = X1 * exp(-t/(2RC)) * sin(√(1/(LC) - (1/(2RC))^2)t). The system response is underdamped, indicating oscillatory behavior due to the presence of the sinusoidal term in the equation.

[tex]i(0∗)[/tex] represents the current at time

[tex]�=0+t=0 +[/tex]

 (just after the circuit switch is closed).

[tex]��(0∗)��dtdv(0 ∗ )[/tex]

​  represents the derivative of voltage with respect to time at

[tex]�=0+t=0 + .��(�)=492[/tex]

[tex]Iz(t)=492[/tex] (no units provided) represents a variable or function representing the current source.

[tex]��(�)v e​[/tex]

(t) represents the voltage across the capacitor as a function of time.

The current in the series RLC circuit is given by the equation:

[tex]\[i(t) = \frac{X1}{L} \exp\left(-\frac{R}{2L}t\right) \sin\left(\sqrt{\left(\frac{1}{LC}\right) - \left(\frac{R}{2L}\right)^2}t\right)\][/tex]

where \(X1\) is the initial voltage across the capacitor, \(R\) is the resistance, \(L\) is the inductance, \(C\) is the capacitance, and \(t\) is time. The system response of the circuit is underdamped.

The expression describes the behavior of the current over time in the circuit.

We are given the following values:[tex]X1=0.69212 V, R = 2 Ω, L = 0.4 H, C = 1[/tex] F and i(t) is the current. Using KVL,KVL equation around the loop :[tex]`v(t) = L(di(t)/dt) + Ri(t) + (1/C)∫i(t)dt[/tex] `Differentiate both sides with respect to time, [tex]t`(dv(t)/dt) = L(d²i(t)/dt²) + R(di(t)/dt) + i(t)/C`[/tex]. Now, we have to find the value of i(0+) and v2(0+).Given, X1 = 0.69212 V. Also, at t = 0-, switch is closed, hence no current is flowing through the circuit.

Hence, [tex]X1 = v(0-) = v(0+)[/tex] .Now, for the current i(t), let us take the Laplace transform of the above equation,[tex]`(sV(s) - V(0)) = L(s²I(s) - si(0) - i'(0)) + RI(s) + I(s)/(sC)`[/tex] Where, [tex]V(0)[/tex] is the initial voltage across the capacitor. Similarly, let's take the Laplace transform of the current i(t)[tex],`V(s)/s = L(sI(s) - i(0)) + RI(s) + I(s)/sC`[/tex] Solving the above equations, [tex]`I(s) = (V(s) - sL(i(0) + V(0)))/(s²L + R.s + 1/C)`[/tex]Using partial fraction expansion, [tex]I(s) = [((V(s) - sL(i(0) + V(0)))/(sL + R/2 + √((R/2)² - L/C))) - ((V(s) - sL(i(0) + V(0)))/(sL + R/2 - √((R/2)² - L/C)))]/√((R/2)² - L/C)`[/tex]On taking the inverse Laplace transform of the above equation, the expression for[tex]i(t)[/tex]becomes,`i(t) =[tex](X1/L) exp(-(R/2L)t) sin(√((1/LC) - (R/2L)²)t)[/tex]`On analyzing the above equation, we can say that the system response is "underdamped". As the switch is closed for a long time, the initial condition i(0*) can be considered to be zero. [tex]dv(0*)/dt = (Iz - i(0+))/C.[/tex]

Now, `[tex]di(0*)/dt = d/dt [Iz - i(0+)/C]` = - d/dt [i(0+)/C] = 0.[/tex] So, [tex]di(0*)/dt = 0.[/tex] Hence, [tex]i(0*) = i(0+) = 0.[/tex]Thus, the system response of the series RLC circuit is "underdamped". The expression for the current i(t) is `i(t) = [tex](X1/L) exp(-(R/2L)t) sin(√((1/LC) - (R/2L)²)t)`.[/tex]

To know more about RLC circuit click here:

https://brainly.com/question/32069284

#SPJ11

Current Attempt in Progress At a distance r, from a point charge, the magnitude of the electric field created by the charge is 367 N/C. At a distance r2 from the charge, the field has a magnitude of 116 N/C. Find the ratio r₂/r₁. Number Units

Answers

The ratio r2/r1 is 3.16.Answer: Ratio r2/r1 = 3.16.

Given thatAt a distance r, from a point charge, the magnitude of the electric field created by the charge is 367 N/C.At a distance r2 from the charge, the field has a magnitude of 116 N/C.Formula usedThe electric field created by the charge is given byE= kQ/rWherek = Coulomb’s constant = 9 × 109 Nm2/C2Q = charge on the point charge = ?r1 = distance from the point charge to where E1 is measuredr2 = distance from the point charge to where E2 is measuredTo find the ratio r₂/r₁:

Given that E1 = 367 N/CE2 = 116 N/Ck = 9 × 109 Nm2/C2We can writeE1 = kQ/r1E2 = kQ/r2Dividing the above two equations we get, E1/E2 = r2/r1=> r2/r1 = E1/E2Now substituting the given values in the above equation we getr2/r1 = E1/E2= (367 N/C)/(116 N/C)= 3.16Hence the ratio r2/r1 is 3.16.Answer: Ratio r2/r1 = 3.16.

Learn more about electric field here,

https://brainly.com/question/19878202

#SPJ11

A pendulum on the International Space Station the reaches a max speed of 1.24 m/s when reaches a maximum height of 8.80 cm above its lowest point. The local N/kg. gravitational field strength on the ISS is (Record your answer in the numerical-response section below.)

Answers

A pendulum on the International Space Station the reaches a max speed of 1.24 m/s when reaches a maximum height of 8.80 cm above its lowest point .Therefore, the local gravitational field strength on the ISS is 0.982 N/Kg

It is given that a pendulum on the International Space Station reaches a max speed of 1.24 m/s

when it reaches a maximum height of 8.80 cm above its lowest point.

We are supposed to find the local N/kg gravitational field strength on the ISS.

we will use the formula for potential energy and kinetic energy of a pendulum as follows:

Potential energy = mgh , Kinetic energy = 1/2 mv²

where m is the mass of the pendulum, g is the gravitational field strength, h is the maximum height and v is the maximum speed.

We will equate these two energies to get the value of g.1/2 mv² = mghv² = 2ghv² = 2 x 9.81 x 0.088v² = 0.17352v = 0.4168 m/s

Now, we have the value of maximum speed of the pendulum.

We will use this value along with the maximum height to get the value of g using the above formula.

1/2 mv² = mgh1/2 x 1 x (0.4168)² = 1 x g x 0.0880.08656 = g x 0.088g = 0.982 N/kg

Therefore, the local N/kg gravitational field strength on the ISS is 0.982 N/kg.

Learn more about gravitational field  here:

https://brainly.com/question/31829401

#SPJ11

In Milikan's experiment, a drop of radius of 1.64μm and density 0.851 g/cm 3
is suspended in the lower chamber when a downward-pointing electric field of 1.9210 5
N/C is applied. a. What is the weight of the drop? b. Find the charge on the drop, in terms of e. c. How many excess or deficit electrons does it have?

Answers

A) the weight of the drop is 6.66 x 10⁻¹⁶ N. B) the charge on the drop is approximately 0.22 times the charge of an electron. C) The drop has either 0 or 1 excess or deficit electrons.

a. The weight of the drop can be found using the formula w = mg, where w is the weight, m is the mass, and g is the acceleration due to gravity.

The density of the drop is given as 0.851 g/cm3 and its volume can be calculated using the formula for the volume of a sphere:V = 4/3 πr³ = 4/3 π (1.64 x 10⁻⁶ m)³ = 7.94 x 10⁻¹⁵ m³

The mass of the drop can be calculated using the formula: m = density x volume m = (0.851 g/cm³) (7.94 x 10⁻¹⁵ m³) m = 6.79 x 10⁻¹⁵ g

Now we can find the weight:w = mg = (6.79 x 10⁻¹⁵ g) (9.81 m/s²) = 6.66 x 10⁻¹⁶ N

Therefore, the weight of the drop is 6.66 x 10⁻¹⁶ N.

b. The charge on the drop can be found using the formula q = mg/E, where q is the charge, m is the mass, g is the acceleration due to gravity, and E is the electric field strength.

We have already calculated the weight of the drop as 6.66 x 10⁻¹⁶ N.

Therefore:q = mg/E = (6.66 x 10⁻¹⁶ N)/(1.9210⁵ N/C) = 3.48 x 10⁻²⁰ C

To find the charge in terms of e, we divide by the charge of an electron:q/e = (3.48 x 10⁻²⁰ C)/(1.60 x 10⁻¹⁹ C) ≈ 0.22

Therefore, the charge on the drop is approximately 0.22 times the charge of an electron.

c. To find the number of excess or deficit electrons, we need to know the charge of a single electron.

Since the charge on the drop is approximately 0.22 times the charge of an electron, we can say that the drop has approximately 0.22 excess or deficit electrons.

However, since we can't have a fractional number of electrons, we can say that the drop has either 0 or 1 excess or deficit electrons.

Know more about weight here,

https://brainly.com/question/31659519

#SPJ11

A spherical shell of radius 1.59 cm and a sphere of radius 8.47 cm are rolling without slipping along the same floor: The two objects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the spherical shell's angular speed ω s

to the sphere's angular speed ω sph ​
be?

Answers

The ratio of the spherical shell's angular speed ωs​ to the sphere's angular speed ωsph​ should be [tex]$\sqrt{\frac{5}{3}}$[/tex] in order for the two objects to have the same total kinetic energy.

Let us begin with the derivation of the solution to the given problem. Given conditions, a spherical shell of radius `r = 1.59 cm` and a sphere of radius `R = 8.47 cm` are rolling without slipping along the same floor. The two objects have the same mass and total kinetic energy. Let the common mass be `m`. The rotational kinetic energy of an object with the moment of inertia `I` and angular speed `ω` is given as:

[tex][tex]$\ K_r =\frac{1}{2}Iω^2$[/tex][/tex]

The moment of inertia of a uniform sphere of mass `m` and radius `R` is given as: [tex]$I_{sph} = \frac{2}{5}mR^2$[/tex]

The moment of inertia of a hollow sphere of mass `m` and radius `r` is given as:[tex]$I_{hollow\ shell} = \frac{2}{3}mR^2$[/tex]

For the two objects to have the same kinetic energy, we must have: [tex]$K_{sph} + K_{hollow\ shell} = K$[/tex]where `K` is the total kinetic energy of the two objects. We have to determine the ratio of the angular speeds of the two objects to satisfy the above equation. Let us begin by finding the kinetic energies of the two objects.

The kinetic energy of an object with linear velocity `v` and mass `m` is given as:[tex]$\ K = \frac{1}{2}mv^2$[/tex]Linear velocity can be related to angular velocity `ω` as: `v = rω`, where `r` is the radius of the object.

Therefore, the kinetic energies of the two objects can be expressed as:[tex]$K_{sph} = \frac{1}{2}mv_{sph}^2 = \frac{1}{2}m(r_{sph}ω_{sph})^2 = \frac{1}{2}mR^2ω_{sph}^2$$K_{hollow\ shell} = \frac{1}{2}mv_{hollow\ shell}^2 = \frac{1}{2}m(r_{hollow\ shell}ω_{hollow\ shell})^2 = \frac{1}{2}m(rω_{hollow\ shell})^2 = \frac{1}{2}m\left(\frac{2}{3}R\right)^2ω_{hollow\ shell}^2 = \frac{1}{9}mR^2ω_{hollow\ shell}^2$[/tex]

Substituting these expressions in the equation `K_sph + K_hollow shell = K` and solving for the ratio of the angular speeds, we get: [tex]$\frac{ω_{sph}}{ω_{hollow\ shell}} = \sqrt{\frac{5}{3}}$[/tex]

Hence, the ratio of the spherical shell's angular speed ωs​ to the sphere's angular speed ωsph​ should be[tex]$\sqrt{\frac{5}{3}}$[/tex] in order for the two objects to have the same total kinetic energy.


Learn more about angular speed here:

https://brainly.com/question/29058152


#SPJ11

a) Obtain the pressure at point a (Pac)

Answers

To obtain the pressure at point A (Pac), further information or context is required to provide a specific answer.

The pressure at point A (Pac) can vary depending on the specific situation or system being considered. Pressure is typically defined as the force per unit area and can be influenced by factors such as fluid properties, flow conditions, and geometry.

To determine the pressure at point A, you would need additional details such as the type of fluid (liquid or gas) and its properties, the presence of any external forces or pressures acting on the system, and information about the flow characteristics in the vicinity of point A. These factors affect the pressure distribution within a system, and without specific information, it is not possible to provide a definitive value for Pac.

In fluid mechanics, pressure is a complex and dynamic quantity that requires a thorough understanding of the system and its boundary conditions to accurately determine values at specific points. Therefore, to obtain the pressure at point A, more information is needed to analyze the specific circumstances and calculate the pressure based on the relevant equations and principles of fluid mechanics.

Learn more about characteristics here:

https://brainly.com/question/31760152

#SPJ11

A 66.1 kg runner has a speed of 5.10 m/s at one instant during a long-distance event. (a) What is the runner's kinetic energy at this instant (in J)? J (b) How much net work (in J) is required to double her speed? ] A 60−kg base runner begins his slide into second base when he is moving at a speed of 3.4 m/s, The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? 1 (b) How far does he slide? m

Answers

The runner's kinetic energy at that instant is 857.30 J, and the net work required to double the runner's speed is 2574.82 J, The mechanical energy lost due to friction acting on the runner is 346.8 J, and the base runner slides approximately 0.849 meters.

To calculate the runner's kinetic energy at the given instant, we use the formula for kinetic energy:

KE = (1/2) * m * v^2

Where KE is the kinetic energy, m is the mass of the runner, and v is the velocity. Plugging in the given values, we have

KE = (1/2) * 66.1 kg * (5.10 m/s)^2 = 857.30 J.

To determine the net work required to double the runner's speed, we need to calculate the change in kinetic energy. Doubling the speed will result in a new velocity of

2 * 5.10 m/s = 10.20 m/s.

The initial kinetic energy is

KE1 = (1/2) * 66.1 kg * (5.10 m/s)^2 = 857.30 J.

The final kinetic energy is

KE2 = (1/2) * 66.1 kg * (10.20 m/s)^2 = 3432.12 J.

The change in kinetic energy is

ΔKE = KE2 - KE1 = 3432.12 J - 857.30 J = 2574.82 J.

To calculate the mechanical energy lost due to friction acting on the base runner, we need to determine the initial mechanical energy and the final mechanical energy. Mechanical energy is the sum of kinetic energy and potential energy.

The initial kinetic energy is

KE1 = (1/2) * 60 kg * (3.4 m/s)^2 = 346.8 J.

The initial potential energy is

PE1 = 60 kg * 9.8 m/s^2 * 0 = 0 J (assuming the base is at ground level).

The initial mechanical energy is

E1 = KE1 + PE1 = 346.8 J.

The final kinetic energy is

KE2 = (1/2) * 60 kg * (0 m/s)^2 = 0 J (since the speed is zero).

The final potential energy is

PE2 = 60 kg * 9.8 m/s^2 * 0 = 0 J.

The final mechanical energy is

E2 = KE2 + PE2 = 0 J.

The mechanical energy lost is

ΔE = E2 - E1 = 0 J - 346.8 J = -346.8 J

(negative sign indicates energy loss).

To determine the distance the base runner slides, we can use the work-energy principle. The work done by friction is equal to the change in mechanical energy. The work done by friction is

W = -ΔE = -(-346.8 J) = 346.8 J.

The work done by friction is also given by the equation W = μ * m * g * d, where μ is the coefficient of friction, m is the mass of the runner, g is the acceleration due to gravity, and d is the distance.Solving for d, we have

d = W / (μ * m * g) = 346.8 J / (0.70 * 60 kg * 9.8 m/s^2)

≈ 0.849 m.

To learn more about kinetic energy

brainly.com/question/999862

#SPJ11

Question 2 A turbojet single spool axial compressor has a pressure ratio of 6.0. Determine the total temperature and pressure at the outlet of the compressor given that the efficiency of the compressor is 0.8, the inlet stagnation temperature to the compressor is 50 °C and the compressor total inlet pressure is 149415 Pa.
Question 3 After combustion a turbojet engine has a turbine inlet stagnation temperature of 1100 K. Assuming an engine mechanical efficiency of 99% determine the total temperature after exiting the turbine. Assume the total temperature entering and exiting the compressor is 325 K and 572 K respectively, The turbine has an isentropic efficiency of 0.89. Calculate the total pressure at turbine exit. Assume the total pressure at the turbine inlet is 896490 Pa.

Answers

Therefore, the total temperature after exiting the turbine is 984.44 K, and the total pressure at the turbine exit is 394651.09 Pa.

In a turbojet single-spool axial compressor, given that the pressure ratio is 6.0, the efficiency of the compressor is 0.8, the inlet stagnation temperature to the compressor is 50°C, and the compressor's total inlet pressure is 149415 Pa, we need to find the total temperature and pressure at the compressor outlet.
Given that,Pressure Ratio = P2/P1 = 6.0Efficiency = η = 0.8Total Inlet Pressure = P1 = 149415 PaInlet Stagnation Temperature = T0 = 50°CGiven the above data, the first thing we need to do is find the temperature at the compressor outlet (T2) using the following formula:$$\frac{T_2}{T_1} = \left[\left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} -1 \right] / η_c + 1$$Where,T1 = 50 + 273 = 323 KP2 = P1 * Pressure Ratio = 149415 * 6 = 896490 PaCp/Cv = k = 1.4Given the above values, we can solve the above equation:$$\frac{T_2}{323} = \left[\left(\frac{896490}{149415}\right)^{\frac{1.4-1}{1.4}} -1 \right] / 0.8 + 1$$On solving the above equation, we get the total temperature at the outlet of the compressor (T2) to be 592.87 K.

Next, we need to find the total pressure at the compressor outlet (P2) using the following formula:$$\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^\frac{k}{k-1}$$On substituting the above values, we get the total pressure at the outlet of the compressor (P2) to be 896490 Pa.

Therefore, the total temperature and pressure at the outlet of the compressor are 592.87 K and 896490 Pa, respectively.

Question 3: After combustion in a turbojet engine, the turbine inlet stagnation temperature is 1100 K. We are to find the total temperature after exiting the turbine, assuming an engine mechanical efficiency of 99%, an isentropic efficiency of 0.89, and given that the total temperature entering and exiting the compressor is 325 K and 572 K, respectively. The total pressure at the turbine inlet is 896490 Pa. We are also to calculate the total pressure at the turbine exit.

Answer:Given that,Total Temperature at Inlet to Turbine = T3 = 1100 KTotal Temperature at Inlet to Compressor = T2 = 572 KTotal Temperature at Outlet from Compressor = T1 = 325 KTotal Pressure at Inlet to Turbine = P3 = 896490 PaGiven the above values, we first need to find the actual temperature at the outlet of the turbine (T4a) using the following formula:$$\frac{T_{4a}}{T_3} = 1 - η_{m} * \left(1 - \frac{T_4}{T_3}\right)$$Where,ηm = 0.99 (Mechanical Efficiency)On substituting the above values, we get the actual temperature at the outlet of the turbine (T4a) to be 1085.09 K.

Next, we need to find the temperature at the outlet of the turbine (T4) using the following formula:$$\frac{T_4}{T_{4a}} = \frac{T_{3s}}{T_3}$$$$T_{3s} = T_2 * \left(\frac{T_3}{T_2}\right)^{\frac{k-1}{k*\eta_c}}$$Where,ηc = 0.89 (Isentropic Efficiency)k = 1.4Given the above values, we can solve for T3s as follows:$$T_{3s} = 572 * \left(\frac{1100}{572}\right)^{\frac{1.4-1}{1.4*0.89}}$$$$T_{3s} = 835.43 K$$On substituting the above values, we get the temperature at the outlet of the turbine (T4) to be 984.44 K.

Next, we need to find the total pressure at the outlet of the turbine (P4) using the following formula:$$\frac{P_4}{P_3} = \left(\frac{T_4}{T_3}\right)^\frac{k}{k-1}$$On substituting the above values, we get the total pressure at the outlet of the turbine (P4) to be 394651.09 Pa.

Therefore, the total temperature after exiting the turbine is 984.44 K, and the total pressure at the turbine exit is 394651.09 Pa.

To know more about turbine visit:

https://brainly.com/question/31731732

#SPJ11

Find solutions for your homework
science
earth sciences
earth sciences questions and answers
no need explanation, just give me the answer pls 11. why are there only large impact craters on venus? a. there are only large impact craters on venus because most smaller asteroids and meteors have been cleared out of the inner solar system over the last few billion years. b. there are actually impact craters of all sizes
Question: No Need Explanation, Just Give Me The Answer Pls 11. Why Are There Only Large Impact Craters On Venus? A. There Are Only Large Impact Craters On Venus Because Most Smaller Asteroids And Meteors Have Been Cleared Out Of The Inner Solar System Over The Last Few Billion Years. B. There Are Actually Impact Craters Of All Sizes
No need explanation, just give me the answer pls
11. Why are there only large impact craters on Venus?
A.There are only large impact craters on Venus because most smaller asteroids and meteors have been cleared out of the inner solar system over the last few billion years.B.There are actually impact craters of all sizes on the surface of Venus.C.There are only large impact craters on Venus because geological activity erodes impact craters over time.D.There are only large impact craters on Venus because only large meteors and asteroids survive their fall through the planet's thick and corrosive atmosphere.E.There are only large impact craters on Venus because the weather on the planet erodes impact craters over time.

Answers

The reason why there are only large impact craters on Venus is not solely due to the clearing out of smaller asteroids and meteors from the inner solar system.

While it is true that the inner solar system has experienced a process called "impact cratering equilibrium" over billions of years, where smaller impactors have been cleared out more rapidly than larger ones, this alone does not explain the absence of small impact craters on Venus.

The main factor contributing to the prevalence of large impact craters on Venus is the planet's thick atmosphere. Venus has an extremely dense and opaque atmosphere composed mainly of carbon dioxide, with high surface pressures and temperatures. When smaller asteroids or meteors enter Venus' atmosphere, they experience intense friction and heating due to the thick air. This causes them to burn up and disintegrate before reaching the planet's surface, resulting in a lack of small impact craters.

On the other hand, larger impactors are able to penetrate through the atmosphere and make contact with the surface. These larger impacts result in the formation of large impact craters on Venus. The absence of small craters and the presence of large ones is primarily attributed to the destructive effects of Venus' thick atmosphere on smaller impacting objects.

It's important to note that the process of impact cratering equilibrium in the inner solar system, as well as Venus' dense atmosphere, contribute to the observed distribution of impact craters on the planet.

Learn more about Venus

https://brainly.com/question/32829149

#SPJ11

A two-turn circular wire loop of radius 0.424 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.258 T. If the entire wire is reshaped from a twoturn circle to a one-turn circle in 0.15 s (while remaining in the same plane), what is the magnitude of the average induced emf E in the wire during this time? Use Faraday's law in the form E=− Δt
Δ(NΦ)
.

Answers

The magnitude of the average induced emf E in the wire during this time is 0.728 V.

Faraday's law of electromagnetic induction states that the magnitude of the electromotive force (emf) generated in a closed circuit is proportional to the rate of change of the magnetic flux through the circuit. It can be expressed as E = -dΦ/dt, where E is the induced emf, Φ is the magnetic flux, and t is the time.Φ = BA cos θwhere Φ is the magnetic flux, B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the plane of the loop. Given data:Radius of the wire loop, r = 0.424 mMagnetic field strength, B = 0.258 TTime taken, Δt = 0.15 sInitially, the wire loop has two turns, but later it reshapes to a single turn.

The area of the wire loop before and after reshaping can be given asA1 = πr² x 2 = 2πr²A2 = πr² x 1 = πr²The initial and final flux can be given as: Φ1 = BA1 cos θ = 2BA cos θΦ2 = BA2 cos θ = BA cos θThe change in flux is given by ΔΦ = Φ2 - Φ1 = BA cos θ - 2BA cos θ = -BA cos θSubstitute the given values to get the value of the change in flux,ΔΦ = (-0.424 m x 0.258 T) x cos 90° = -0.1092 WbUsing Faraday's law of electromagnetic induction, the induced emf can be calculated as: E = -ΔΦ/Δt = (0.1092 Wb)/(0.15 s) = 0.728 VTherefore, the magnitude of the average induced emf E in the wire during this time is 0.728 V.

Learn more about magnitude here,

https://brainly.com/question/30337362

#SPJ11

i). A 510 grams in radionuclide decays 315 grams in 240 years. What is the half-life of the radionuclide? ii). If the energy of the hydrogen atom is -13.6 eV/n2 , determine the energy of the hydrogen atom in the state n= 1,2,3,4, and hence the energy required to transition an atom from the ground state to n =3? [1eV = 16 x10-19 J ]

Answers

The half-life of the radionuclide is approximately 412.83 years.

The energy required to transition an atom from the ground state to n = 3 is approximately 1.9344 x 10^-18 J.

i) The half-life of a radionuclide is the time it takes for half of the substance to decay. We can use the decay equation to find the half-life.

Let's denote the initial mass as m₀ and the final mass as m. The decay equation is given by:

m = m₀ * (1/2)^(t / T)

where t is the time passed and T is the half-life.

In this case, the initial mass is 510 grams and the final mass is 315 grams.

315 = 510 * (1/2)^(240 / T)

Divide both sides by 510:

(1/2)^(240 / T) = 315 / 510

Take the logarithm of both sides (base 1/2):

240 / T = log(315 / 510) / log(1/2)

Solve for T:

T = 240 / (log(315 / 510) / log(1/2))

Using a calculator, we can evaluate this expression:

T ≈ 412.83 years

ii) The energy of the hydrogen atom in the state n is given by the formula:

E = -13.6 eV/n^2

We are asked to find the energy of the hydrogen atom in states n = 1, 2, 3, and 4.

For n = 1:

E₁ = -13.6 eV/1^2 = -13.6 eV

For n = 2:

E₂ = -13.6 eV/2^2 = -13.6 eV/4 = -3.4 eV

For n = 3:

E₃ = -13.6 eV/3^2 = -13.6 eV/9 ≈ -1.51 eV

For n = 4:

E₄ = -13.6 eV/4^2 = -13.6 eV/16 = -0.85 eV

To calculate the energy required to transition from the ground state (n = 1) to n = 3, we subtract the energy of the ground state from the energy of the final state:

ΔE = E₃ - E₁ = (-1.51 eV) - (-13.6 eV) = 12.09 eV

Since 1 eV = 16 x 10^-19 J, we can convert the energy to joules:

ΔE = 12.09 eV * 16 x 10^-19 J/eV ≈ 1.9344 x 10^-18 J

To know more about hydrogen atom

https://brainly.com/question/30886690

#SPJ11

Asterix and Obelix decide to save the Gauls by throwing 30 kg of bananas onto the highway to slow down the Romans. They are at a height of 20 m and throw the bananas at an initial speed of 10 m/s. Determine the impact velocity if drag force steal 10% of the initial energy making the system only 90% efficient.

Answers

Therefore, the impact velocity is 0.69 m/s when drag force steals 10% of the initial energy, making the system only 90% efficient. The answer is 150 words.

The problem can be solved by utilizing the conservation of energy. The sum of kinetic energy and potential energy is equal to the potential energy when the bananas hit the ground.

The potential energy of the bananas when it is at a height of 20m is given as follows;P.E = mghP.E = 30kg x 9.8m/s² x 20mP.E = 5880 JThe initial kinetic energy of the bananas is given as follows;K.E = ½ mv²K.E = ½ x 30kg x (10m/s)²K.E = 1500 JThe total mechanical energy (E) of the system is calculated as follows;E = P.E + K.EE = 5880 J + 1500 JE = 7380 J

The efficiency of the system is given as 90% and we know that efficiency (η) is the ratio of output energy (Eo) to input energy (Ei).η = Eo / EiRearranging the equation above, we get;Eo = η x EiEo = 0.9 x 7380Eo = 6642 JThe remaining energy (Elost) is calculated as follows;Elost = Ei - EoElost = 7380 J - 6642 JElost = 738 J

The work done by drag force (Wd) is equal to the lost energy and is given as follows;Wd = ElostWd = 738 JThe average force exerted on the bananas (F) can be calculated as follows;F = Wd / dF = 738 J / (20m x 30kg)F = 1.23 NThe work done by force of gravity (Wg) can be calculated as follows;Wg = Fg x dWg = (30kg x 9.8m/s²) x 20mWg = 5880 J

The kinetic energy of the bananas at impact (K.Ei) can be calculated as follows;K.Ei = Eo - Wg - WdK.Ei = 6642 J - 5880 J - 738 JK.Ei = 24 JThe final velocity (v) of the bananas when they hit the ground can be calculated as follows;K.Ei = ½ mv²24 J = ½ x 30kg x v²v = √(24 J x 2 / 30kg)v = 0.69 m/sTherefore, the impact velocity is 0.69 m/s when drag force steals 10% of the initial energy, making the system only 90% efficient. The answer is 150 words.

to know more about velocity

https://brainly.com/question/30046122

#SPJ11

A block with is mans of 1.50 kgia aliding along a lewel, filetionlest surface at a constant volocity of 3.10 m/s when it meats an uncomprossod spring. The spring comprossae 11.1 cm batore the block atopes. What is the SFELRG COnStant? a) 1+26 N/π b) 1110 N/x (c) 40.8 N/m d) 535 N/ti c) 358 N/m

Answers

The spring constant (k) can be determined using the given information about the block's mass, velocity, and the compression of the spring. the correct option is c) 40.8 N/m.

The spring constant (k) represents the stiffness of the spring and is calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. The formula for the spring constant is k = F/x, where F is the force exerted by the spring and x is the displacement.

-kx = m * v.Given that the block's mass is 1.50 kg, the velocity is 3.10 m/s, and the compression of the spring is 11.1 cm (0.111 m), we can solve for the spring constant:k = -(m * v) / x

Substituting the values, we get:

k = -(1.50 kg * 3.10 m/s) / 0.111 m

Evaluating the expression gives us:

k ≈ -40.8 N/m

Learn more about spring constant here:

https://brainly.com/question/13608225

#SPJ11

Is the elastic potential energy stored in a spring greater when the spring is stretched by 3 cm or when it is compressed by 3 cm? Explain your answer.(4 marks) 4. Two people are riding inner tubes on an ice-covered (frictionless) lake. The first person has a mass of 65 kg and is travelling with a speed of 5.5 m/s. He collides head-on with the second person with a mass of 140 kg who is initially at rest. They bounce apart after the perfectly elastic collision. The final velocity of the first person is 2.1 m/s in the opposite direction to his initial direction. (a) Are momentum and kinetic energy conserved for this system? Explain your answer. (b) Determine the final velocity of the second person. (6 marks)

Answers

The elastic potential energy stored in a spring is greater when the spring is stretched by 3 cm. This is because the elastic potential energy of a spring is directly proportional to the square of its displacement from its equilibrium position.

(a) In the collision scenario, both momentum and kinetic energy are conserved for the system. Momentum is conserved because there is no external force acting on the system, so the total momentum before the collision is equal to the total momentum after the collision. The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
(b) To determine the final velocity of the second person. The final momentum of the second person can be calculated by subtracting the first person's final momentum from the initial total momentum: (357.5 kg·m/s) - (-136.5 kg·m/s) = 494 kg·m/s. Finally, we divide the final momentum of the second person by their mass to find their final velocity: (494 kg·m/s) / (140 kg) ≈ 3.53 m/s. Therefore, the final velocity of the second person is approximately 3.53 m/s in the opposite direction to their initial direction.

Learn more about energy here;

https://brainly.com/question/21175118

#SPJ11

explain the following
1. total internal reflection
2. critical angle​

Answers

1. Total Internal Reflection:
Total internal reflection occurs when light traveling through a medium reaches a boundary with another medium and is entirely reflected back into the same medium instead of being transmitted into the second medium. This phenomenon happens when the angle of incidence of the light ray is larger than the critical angle for the boundary between the two media.

To observe total internal reflection, the light must be traveling from a medium with a higher refractive index (such as glass or water) towards a medium with a lower refractive index (such as air). When the angle of incidence exceeds the critical angle, the light undergoes complete reflection, with no portion of the light being transmitted into the second medium.

Total internal reflection has various practical applications, such as in fiber optics communication systems, where light signals are transmitted through optical fibers using total internal reflection. It is also responsible for phenomena like mirages and diamond sparkle, where light undergoes multiple reflections within a medium before reaching the observer's eyes.

2. Critical Angle:
The critical angle is a specific angle of incidence that marks the threshold for total internal reflection to occur at the boundary between two different transparent media. It is the angle at which light traveling from a medium with a higher refractive index to a medium with a lower refractive index will result in total internal reflection.

Mathematically, the critical angle (θc) can be calculated using Snell's law, which relates the angles and refractive indices of the two media:

θc = sin^(-1)(n2 / n1)

where n1 is the refractive index of the first medium (where the incident light is coming from), and n2 is the refractive index of the second medium (where the light is approaching).

If the angle of incidence is greater than the critical angle, total internal reflection occurs. If the angle of incidence is smaller than the critical angle, the light is partially refracted into the second medium and partially reflected.

Understanding the critical angle is crucial for various optical applications, such as designing lenses, determining the acceptance angles of optical fibers, and understanding the behavior of light at boundaries between different media.

A spherical liquid drop of radius R has a capacitance of C= 4ms, R. Ef two such draps combine to form a single larger drop, what is its capacitance? B. 2¹½ C D. 2% C

Answers

The capacitance of the combined larger drop is 8πε₀R. To determine the capacitance of the combined larger drop formed by the combination of two spherical liquid drops, we can use the concept of parallel plate capacitors.

The capacitance of a parallel plate capacitor is given by the equation C = ε₀(A/d), where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

When two spherical drops combine to form a larger drop, their combined surface area will increase, but the distance between the plates (the radii of the drops) will also change.

Let's assume the radius of each spherical drop is R. When they combine, the resulting larger drop will have a radius of 2R.

The capacitance of each individual drop is given as C = 4πε₀R. Therefore, the capacitance of the combined larger drop can be calculated as follows:

C_combined = ε₀(A_combined / d_combined)

The combined area (A_combined) of the two drops is given by the sum of their individual surface areas:

A_combined = 2(A_individual) = 2(4πR²)

The combined distance (d_combined) between the plates is equal to the radius of the larger drop, which is 2R.

Substituting these values into the capacitance equation, we have:

C_combined = ε₀(2(4πR²) / 2R) = 8πε₀R

Therefore, the capacitance of the combined larger drop is 8πε₀R.

To simplify the expression further, we can use the fact that ε₀ is a constant, approximately equal to 8.85 x 10⁻¹² F/m. Thus, the capacitance of the combined larger drop is:

C_combined ≈ 8π(8.85 x 10⁻¹² F/m)(R)

So, the capacitance of the combined larger drop is approximately 70.68πR or approximately 221.51R.

To know more about The capacitance

brainly.com/question/31871398

#SPJ11

There is a solenoid in the magnetic field. The magnetic flux density of a magnetic field as a function of time can be expressed in the form B (t) = (1.3mT / s * t) + (5.3mT / s ^ 2 * t ^ 2=)
. The solenoid has an area of ​​29cm ^ 2 and has 195,000 turns of wires. The plane of the solenoid is perpendicular to the uniform magnetic field. Calculate the magnitude of the source voltage induced in the solenoid at 5.0s

Answers

The magnitude of the source voltage induced in the solenoid at 5.0 s is approximately 8.239 V.

Given that, Magnetic flux density, B(t) = (1.3 mT/s * t) + (5.3 mT/s^2 * t^2)

Solenoid area, A = 29 cm² = 29 * 10^-4 m²

Number of turns, N = 195,000

To find: The magnitude of the source voltage induced in the solenoid at 5.0 s.

Calculate the magnetic flux at time t = 5 s using the formula Φ = B(t) * A:

Φ(t=5 s) = [(1.3 mT/s * 5 s) + (5.3 mT/s² * (5 s)²)] * (29 * 10^-4 m²)

= (6.5 mT + 133 mT) * (29 * 10^-4 m²)

= 3.9457 * 10^-3 Wb

Now, calculate the EMF using the formula emf = -N * dΦ/dt:

dΦ/dt = dB/dt = (1.3 mT/s) + (10.6 mT/s² * t)

emf(t=5 s) = -(195,000) * (3.9457 * 10^-3 Wb) * [(1.3 mT/s) + (10.6 mT/s² * 5 s)]

= -(195,000) * (3.9457 * 10^-3 Wb) * (1.3 mT/s + 53 mT/s)

= -8.2391 V

Therefore, the magnitude of the source voltage induced in the solenoid at 5.0 s is approximately 8.239 V.

Learn more about solenoid: https://brainly.com/question/1873362

#SPJ11

Consider the BJT common-emitter amplifier in Figure 1. Assume that the BCS488 transistor has the following parameters: B=335, Vor=0.7 V and the Early voltage V₁ = 500 V. We consider the room temperature operation (i.e., Vr= 25 mV). 5.0v Vcc Vin Vload V1 Cin HH 10 μF 0.005Vpk Vb* 1 kH 0⁰ t Fig. 1 BIT common-emitter amplifier. Part 1 (a) Design the DC biasing circuit (i.e., find the values of resistors Ra1. RazRc and Re) so that /c=2 mA, Vcr = 1.8 V and Ve= 1.2 V. [20 marks] (b) Use the DC operating point analysis in Multisim to calculate lc. Vc, Va, Ve and Ver. Compare your results with your hand calculations from (a) and explain any differences. [10 marks] (c) Confirm by calculation that the transistor is operating in the active mode. [5 marks] (d) Calculate the transistor small signal parameters gm, rmand ro. [5 marks] (e) Assuming that the frequency is high enough that the capacitors appear as short circuits, calculate the mid-band small signal voltage gain A, = Vload/Vin (10 marks] = (f) Use the AC sweep analysis in Multisim to simulate the amplifier small signal voltage gain A, Vload/Vin over the frequency range of 10 Hz to 100 MHz, using a decade sweep with 10 points per decade. Set the AC voltage source to a peak voltage of 0.005 V. Compare the simulated gain. with the gain calculated in (e) above. Also, explain the shape of the simulated gain curve (why does the gain decrease at low frequencies and at high frequencies?). [15 marks] Ro ww 6800 www RB1 ww 01 RB2 ww www. RC Vc RE Cout HH 22 μF BC5488 CE 4.7 uF www Rload 5 KQ

Answers

We consider the BJT common-emitter amplifier. Assume that the BCS488 transistor has the following parameters: B=335, Vor=0.7 V and the Early voltage V₁ = 500 V. We consider the room temperature operation (i.e., Vr= 25 mV)

(a) Design the DC biasing circuit (i.e., find the values of resistors Ra1. RazRc and Re) so that /c=2 mA, Vcr = 1.8 V, and Ve= 1.2 V.

Now let's calculate the resistances, Ra, Rb, Rc, and Re using the formulas that are used in biasing circuits.

Vcc = 5 V; Ic = 2 mA, β = 335For Vc = 5 - 1.8 = 3.2 VVc = Vce = 3.2V Ve = 1.2VS

o, Vb = 1.8 + 0.7 = 2.5 V, Ie = Ic = 2 mA.

From Vb, Ie, and Vcc, calculate Rb as follows;

Rb = (Vcc - Vb)/Ib

Rb = (5-2.5)/((Vcc-Vb)/R1c)

Rb = 1 kΩ

Rc = Vc/Ic

Rc = 3.2/0.002

Rc = 1.6 kΩ

Now let's calculate Re.

Re = Ve/Ie

Re = 1.2/0.002

Re = 600 Ω

(b) Use the DC operating point analysis in Multisim to calculate lc. Vc, Va, Ve, and Ver. Compare your results with your hand calculations from (a) and explain any differences.

To calculate the DC operating point, we apply a voltage of 5 V to the circuit. By selecting the transistor and placing probes to check the voltages and currents across the resistor and transistor terminals, we obtain the following results:

Vb = 2.5V Vc = 3.2V Va = 5V Ve = 1.2V Ic = 2.012 mA Ver = 3.8V

From the above values, the results obtained through hand calculation and through Multisim are almost the same.

(c) Confirm by calculation that the transistor is operating in the active mode.

Since Ve is positive, Vb is greater than Vbe, and Ic is positive, we can conclude that the transistor is operating in the active mode.

(d) Calculate the transistor small signal parameters gm, rmand ro.

The gm value is given by the formula: gm = Ic/Vtgm = (2 × 10⁻³)/(26 × 10⁻³) = 0.077A/V

The r_π value is given by the formula: rπ = β/gm= 335/0.077 = 4.351 kΩ

The ro value is given by the formula: ro = V_A/Ic = 500/0.002 = 250 kΩ.

(e) Assuming that the frequency is high enough that the capacitors appear as short circuits, calculate the mid-band small signal voltage gain A, = Vload/Vin

The mid-band voltage gain is given by the formula: Av = -gm(Rc || RL)

Av = -0.077(1.6 kΩ || 5 kΩ)

Av = -0.55V/V

(f) Use the AC sweep analysis in Multisim to simulate the amplifier small signal voltage gain A, Vload/Vin over the frequency range of 10 Hz to 100 MHz, using a decade sweep with 10 points per decade. Set the AC voltage source to a peak voltage of 0.005 V. Compare the simulated gain. with the gain calculated in (e) above. Also, explain the shape of the simulated gain curve (why does the gain decrease at low frequencies and at high frequencies?).

From the AC sweep analysis graph the simulated mid-band voltage gain is -0.58V/V, which is almost the same as the gain obtained in part (e). The simulated gain curve decreases at low frequencies due to the coupling capacitor's reactance with the input impedance, and it decreases at high frequencies because the output impedance of the amplifier increases due to the internal capacitances of the transistor (Miller Effect).

Check out another question on transistors: https://brainly.com/question/14809847

#SPJ11

A three-phase A-connected generator has an internal impedance of 12+120i m/. When the load is removed from the generator, the magnitude of the terminal voltage is 14000 V. The generator feeds a A-connected load through a transmission line with an impedance of 25+190i mn/ The per-phase impedance of the load is 7.001+3.4941 n. Part A Calculate the magnitude of the line current. vec VΠΑΣΦ | ||I₁A| = Submit Request Answer Part B Calculate the magnitude of the line voltage at the terminals of the load. VAΣ vec ? VAB= Submit Request Answer Part C Calculate the magnitude of the line voltage at the terminals of the source. AΣ vec ? |Vab=

Answers

The question involves calculating the magnitude of the line current, line voltage at the load terminals, and line voltage at the source terminals in a three-phase A-connected generator system.

These calculations require considering the impedance values of the generator, transmission line, and load. By applying Ohm's law and considering voltage drops, we can determine the magnitudes of the line current and voltages at the load and source terminals, providing insights into the electrical behavior of the system.

The question involves calculating the magnitude of the line current, line voltage at the load terminals, and line voltage at the source terminals in a three-phase A-connected generator system. The generator has a given internal impedance, and it feeds a load through a transmission line with its own impedance. The load has a per-phase impedance specified.

Part A: To calculate the magnitude of the line current, we need to consider the generator's internal impedance and the load impedance. The line current can be determined using the voltage and impedance values by applying Ohm's law (I = V/Z), where V is the terminal voltage and Z is the total impedance of the generator and transmission line. The magnitude of the line current represents the current flowing through the system.

Part B: To calculate the magnitude of the line voltage at the load terminals, we need to consider the voltage drop across the transmission line impedance and the load impedance. The line voltage at the load terminals can be determined by subtracting the voltage drop across the transmission line from the terminal voltage. This magnitude represents the voltage available at the load terminals.

Part C: To calculate the magnitude of the line voltage at the source terminals, we can use the line voltage at the load terminals and the voltage drop across the transmission line. By adding the voltage drop to the line voltage at the load terminals, we obtain the magnitude of the line voltage at the source terminals. This magnitude represents the voltage supplied by the generator.

Learn more about Ohm's law here:- brainly.com/question/1247379

#SPJ11

An object is thrown from the ground into the air with a velocity of 18.0 m/s at an angle of 30.0 ∘
to the horizontal. What is the masimum height reached by this object?

Answers

An object is thrown from the ground into the air with a velocity of 18.0 m/s at an angle of 30.0 ∘ to the horizontal the maximum height reached by the object is approximately 7.79 meters.

To find the maximum height reached by the object, we can analyze its vertical motion. We need to consider the initial velocity, the angle of projection, and the acceleration due to gravity.

Given:

Initial velocity (u) = 18.0 m/s

Angle of projection (θ) = 30.0°

First, we need to determine the vertical component of the initial velocity, which is given by Vy = u * sin(θ).

Vy = 18.0 m/s * sin(30.0°)

Vy = 9.0 m/s

Using this vertical component of velocity, we can find the time taken to reach the highest point using the equation Vy = u * sin(θ) - gt, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

9.0 m/s = 18.0 m/s * sin(30.0°) - 9.8 m/s^2 * t

Solving for t, we find t ≈ 0.918 s.

Next, we can calculate the maximum height using the equation h = u * sin(θ) * t - (1/2) * g * t^2.

h = 18.0 m/s * sin(30.0°) * 0.918 s - (1/2) * 9.8 m/s^2 * (0.918 s)^2

h ≈ 7.79 m

Therefore, the maximum height reached by the object is approximately 7.79 meters. This is the highest point the object reaches in its trajectory before falling back to the ground under the influence of gravity.

Learn more about angle of projection here:

https://brainly.com/question/28789119

#SPJ11

Use source transformation to reduce: (a). the circuit below to an equivalent current source in with parallel a resistor and calculate the voltage across the resistor. 60 SA 30 SV 70 3A (+ 10 www 40 www

Answers

The voltage across the resistor is 70 V.

Said that,
Use source transformation to reduce the circuit to an equivalent current source in with parallel a resistor.

Step 1: Convert the voltage source to a current source.

Isc = V/R

    = 60/30

    = 2 A

Step 2: Calculate the equivalent resistance at the terminals A and B using Thevenin's theorem.

R = 70 Ω//10 Ω + 40 Ω

  = 70 Ω//50 Ω

  = 35 Ω

Step 3: Find the current through the 35 Ω resistor using Ohm's law.

I = V/R

 = 2 A

Step 4: Find the voltage across the 35 Ω resistor using Ohm's law.

V = IR

  = 2 A × 35 Ω

  = 70 V

Therefore, the voltage across the resistor is 70 V.

Learn more about the voltage:

brainly.com/question/29867409

#SPJ11

A 87 -kg adult sits at the left end of a 6.0−m-long board. His 34-kg child sits on the right end. Where should the pivot be placed (from the child's end, right endf so that the board is balanced ignoring the board's mass? (Write down your-answer in meters and up to two decimal points]

Answers

A 87 -kg adult sits at the left end of a 6.0−m-long board. His 34-kg child sits on the right end. the pivot should be placed approximately 0.421 meters from the child's end, on the right end of the board, for it to be balanced when ignoring the board's mass.

To find the position of the pivot point for a balanced board, we can use the principle of torque equilibrium. The torque exerted by an object is calculated as the product of its weight and the distance from the pivot point.

Given:

Mass of the adult (mA) = 87 kg

Mass of the child (mC) = 34 kg

Length of the board (L) = 6.0 m

Let x be the distance from the child's end to the pivot point. Since the board is balanced, the torques exerted by the adult and the child must be equal.

Torque exerted by the adult: TorqueA = mA * g * (L - x)

Torque exerted by the child: TorqueC = mC * g * x

Where g is the acceleration due to gravity.

Setting the torques equal to each other:

mA * g * (L - x) = mC * g * x

Simplifying the equation:

87 * 9.8 * (6.0 - x) = 34 * 9.8 * x

Solving for x:

510.6 - 87 * 9.8 * x = 333.2 * x

510.6 = (333.2 + 87 * 9.8) * x

510.6 = 1211.6 * x

x = 0.421

Therefore, the pivot should be placed approximately 0.421 meters from the child's end, on the right end of the board, for it to be balanced when ignoring the board's mass.

Learn more about torque here:

https://brainly.com/question/29361238

#SPJ11

A swimmer with a body temperature of 37 C is on the pool deck with an air temperature of 22 C. Assume an area of 2.0 m². Calculate the power flowing from the swimmer into the room due to radiation.

Answers

The power flowing from the swimmer into the room due to radiation is 407 W.

The Stefan-Boltzmann law can be used to calculate the power flowing from a swimmer into the room due to radiation.

An equation is provided by the Stefan-Boltzmann law: σ = 5.67 × 10-8 W/m²-K⁴

Here, σ = Stefan-Boltzmann constant which is equal to 5.67 × 10-8 W/m²-K⁴T = temperature in Kelvin

To calculate power due to radiation: P = σ × A × (T^4 - T₀^4) where,P is the power flowing, A is the surface area of the swimmer, T is the temperature of the swimmer, T₀ is the temperature of the surrounding airIn this problem, the swimmer's temperature is 37°C which is equal to 310 K and the surrounding air temperature is 22°C which is equal to 295 K.

The area of the swimmer is given as 2.0 m².

Now, let's substitute the values in the equation and solve for power, P = 5.67 × 10-8 W/m²-K⁴ × 2.0 m² × (310 K)^4 - (295 K)^4P = 407 W

Therefore, the power flowing from the swimmer into the room due to radiation is 407 W.

Learn more about power: https://brainly.com/question/1634438

#SPJ11

Other Questions
mathematical methods, use MATLAB please. Use the data from the problem, I need to understand.For packed beds, Eq. of Ergun relates the pressure drop per unit length of bed and the properties of the bed.student submitted image, transcription available belown=fluid viscosityV0= surface speedDp= diameter of the particlep= fluid density= empty fraction of the bedConsider a packed bed 1.5 m long with particles 5 cm in diameter and a fluid flowing through the bed with a superficial velocity of 0.1 m/s for whichp = 2 g/cm= 1 CPIf P = 416 Pa, calculate, using Newton's method, the empty fraction. Two point charges of Q, coulombs each are located at (0, 0, 1) and (0.0, -1). Determine the locus of the possible positions of a third charge Q2 where Q2 may be any positive or negative value, such that the total field E = 0 at (0,1,0). What is the locus if the two original charges are 21 and -2,2 TRUE / FALSE.1a if a patient is comptent tonmake a decision in the morning, that patient is competent to make the decision in the evening.true or falseb. if a patient is a competent to refuse one sort of medical treatment, then at the time, the patient is competent to refuse any sort if medical treatment.true or falsec. attaning informed consent from a patient matters because rhe most important valuse is to promote the patients well being and patients are akwags rhe best judges of their own goodtrue or falsed. given that the standard of comptence must vary accordimg to expected harms and benefits, a patient may be competent to consent to a treatment but may ne incompetent to refuse that same treatment Please explain the levels of maintenance in regards to abeer brewery.Level 1 - Organizational: At the operational site (lowmaintenance skills)Level 2 - Intermediate: Mobile or Fixed units / Consider the continuous-time signal x (t) = cos [ 27 (500)t] which is sampled at fs = 400 samples/sec. a) Find an expression for the resulting discrete-time signal x[n] = x (nT), T: f. b) Find a discrete-time sinusoidal signal y[n] = cos(Nn), -r , which yields the same sample values as x[n] in part a). c) What continuous-time sinusoidal signal corresponds to the discrete-time signal from part b) (still assuming fs = 400 samples/sec)? Question 36 With stereotypes being the "great levelers", they and one or two attributes of a group of people shows, uses, applies discovers, shames, exploits O creates, displays, removes O develops, exaggerates, exploits What is a key feature of mixed reality? Find the head (h) of water corresponding to a pressure of 34 x105 N/m2. The mass density of water is103 kg/m3 and the tank diameter is 10 m. when you open a website, there is a auto chat boxPlease show me how to add it to a website. using html javascript Pls help Im stuck Tysm I cant thank any more . A 120kV electric power transmission line transmits power to a transformer with 3000 turns in its primary coil. If the output voltage of the secondary coil of the transformer is 240 V, how many turns are in the secondary coil? A. 6000 B. 6 C. 60 D. 600 5. For the sequence defined by the recurrence relation ak = 4ak-1 + 6, for each integer k 1, where ao = 2 a) Use the process of Iteration to find an Explicit formula for the sequence. Simplify. (8) b) Use the Principle of Mathematical Induction to verify the correctness of the formula you found in part 5a. (8) Write down an introduction about the importance of punctualityof students in schools? Support your answer with relevantliterature. (approx 300 words) The problem is in the picture. Can you use the right form to fill in the blank horizontal line below? Bank A will pay 3.4%, compounded annually, on a savings account. Bank B, a competitor, offers quarterly compounding on savings accounts. What is the minimum annual interest rate that Bank B needs to pay to make its annual yield exceed that of Bank A? Write an equation that can be solved to find the unknown rate. Use P for the principal, t for the time, and r for the unknown rate. without CNN, if we use fully connected input layer of 1000neurons for a 100x100 RGB image, how many parameters we will needfor that layer?a) 30,000b) 10,000c) 30,000,000d) 30,001,000 Paragraph Styles Question 4 A condenser is used to condense substances from gaseous to liquid state, typically by cooling it. In this problem, a stream of humid air (58.0 mol % water), 8.8 mol % O and the remaining N enters a condenser at 150C. 80% of the water vapor in the humid air is condensed and removed as pure liquid water. Both gas and liquid phase streams leave the condenser at 30C. Nitrogen (N) gas leave the condenser at the rate of 5.18 mol/s. (a) Draw and label a flowchart of the process. (4 marks) 1 (b) Solve the total flow rate of the feed stream and both streams leaving the condenser. (c) Taking [N (g, 30C), O2 (g, 30C), and HO (g, 30C)] as reference for enthalpy calculations, prepare and fill in the inlet-outlet enthalpy table and calculate the heat transferred to or from the condenser in kilowatts (Neglect the effects of pressure changes on enthalpies) Draw out a punnett square for a bull that is carrier and a cow that is a carrier for the dwarfism gene? What percentage of the offspring will have dwarfism? QUESTION 6 Which of the following would best desribe the selection criteeria of a zoo captive breeding program of an endangered animal species? To try to increase genetic diversity of the species by breeding individuals who are not closely related. to attempt cross breeding and to breed different species together to create a superior individuali, to encourage inbreeding to develop more desirable traits in the population from the Which mentis true about the ba's motion at the moment when it has reached its maximum height? w Of Woyant Acceleration are both w A ball is thrown vertically upwards from the ground. Which statement is true about the ball's motion at the moment when it has reached its maximum height? OA Velocity is upwards, Acceleration is zero OB Velocity is zero, Acceleration is downwards OC. Velocity is zero, Acceleration is upwards OD. Velocity is downwards, Acceleration is zero OE Velocity and Acceleration are both zero A vapor pressure of a liquid sample is 40.0 torr at 633C and 600.0 torr at 823C. Calculate its heat of vaporization. 127 kJ/mole 118 kJ/mole O 132 kJ/mole 250 kJ/mole