1. The ratio of the speed of light in air to the speed of light in the water, n = 1/1.33 = 0.7518. 2. Hence, the angle of refraction is `48.76°`.3. Therefore, the distance from the edge of the pool where the light hits the bottom of the pool is 8.1 + 2.491 = 10.59 m.4. The location of the image is `-40/3 cm`. 5. Therefore, the image is virtual and erect.6.Therefore, the height of the image is `-1.25 cm`.
Part 1: The angle of incidence is given by sin i/n = sin r, where i is the angle of incidence, r is the angle of refraction, and n is the refractive index.
sin i = 1.7/8.1 = 0.2098.
n is the ratio of the speed of light in air to the speed of light in the water, n = 1/1.33 = 0.7518.
Therefore, sin r = sin i/n = 0.2796. Hence, r = 16.47. Therefore, the angle of incidence is `73.53°`.
Part 2: The angle of incidence is given by sin i/n = sin r, where i is the angle of incidence, r is the angle of refraction, and n is the refractive index.
The angle of incidence is 90° since the light ray is travelling perpendicular to the surface of the water.
The refractive index of water is 1.33, hence sin r = sin(90°)/1.33 = 0.7518`.
Therefore, r = 48.76°.
Hence, the angle of refraction is `48.76°`.
Part 3: Using Snell's Law, `n1*sin i1 = n2*sin i2, where n1 is the refractive index of the medium where the light ray is coming from, n2 is the refractive index of the medium where the light ray is going to, i1 is the angle of incidence, and `i2` is the angle of refraction. In this case, `n1 = 1.00`, `n2 = 1.33`, `i1 = 73.53°`, and `i2 = 48.76°`.
Therefore, `sin i2 = (n1/n2)*sin i1 = (1/1.33)*sin 73.53° = 0.5011`.The distance from Diana to the edge of the pool is `8.1 - 1.7*tan 73.53° = 2.428 m.
Hence, the distance from the edge of the pool to the point where the light ray hits the bottom of the pool is `2.428/tan 48.76° = 2.491 m.
Therefore, the distance from the edge of the pool where the light hits the bottom of the pool is 8.1 + 2.491 = 10.59 m.
Part 4: Calculate the location of the image, in cm
Using the lens formula, 1/f = 1/v - 1/u , where f is the focal length of the mirror, u is the object distance and v is the image distance, we have:`1/f = 1/v - 1/u => 1/(-10) = 1/v - 1/4 => v = -40/3 cm.
The location of the image is `-40/3 cm`
Part 5:Since the object distance `u` is positive, the object is in front of the mirror. Since the image distance `v` is negative, the image is behind the mirror.
Therefore, the image is virtual and erect.
Part 6: Calculate the height of the image, in cm
The magnification m is given by m = v/u = -10/4 = -2.5`.The height of the image is given by h' = m*h`, where `h` is the height of the object. Since the height of the object is 0.500 cm, the height of the image is `h' = -2.5*0.500 = -1.25 cm.
Therefore, the height of the image is `-1.25 cm`.
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Here is a graph of my dog walking in my yard. a. What is the dog's displacement after 5 s ? b. What is the dog's distance travelled after 5 s ? c. At what position (if any) is the dog stopped? d. What is the dog's velocity at t=4 s ?
a. The dog's displacement after 5 seconds is equal to the change in position. To determine this, we must determine the distance between the final position and the initial position.
The dog's initial position was zero, and its final position was 1 meter west (negative direction), so the displacement is equal to 1 meter in the negative direction.
Displacement = final position - initial position = -1 m - 0 m = -1 m.
b.
The distance traveled is the total distance covered by the dog. We must determine the sum of the magnitudes of each vector quantity in this case. The displacement from the previous part was equal to 1 m, but we must now account for the distance that the dog covered in the positive direction (east) before moving back west. 2 m + 1 m = 3 m total distance covered. The dog's distance traveled after 5 seconds is equal to 3 meters.
c. The dog is motionless when its position remains constant. The dog is stationary between 2 and 3 seconds because the graph is flat. The dog is not in any position when it is stopped.
d.
Velocity is defined as the rate at which the position changes over time. If the position increases over time, the velocity is positive, whereas if the position decreases over time, the velocity is negative.
When the position remains constant, the velocity is zero. The graph is flat between 2 and 3 seconds, so the velocity is zero. When the dog is at a position of 1 meter west of the origin at 5 seconds, the dog's velocity is calculated as follows:
Velocity = displacement/time = (-1 m - 0 m) / (5 s - 0 s) = -1/5 m/s.
The dog's velocity at t = 4 s is -1/5 m/s.
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2.17 Compute and plot the solar irradiance at the top of the earth's atmosphere emitted from temperatures of 5000,5500 , and 6000 K. Compare your results with those presented in Figs. 2.9 and 2.10.
The
solar irradiance
emitted from temperatures of 5000 K, 5500 K, and 6000 K at the top of the earth's atmosphere can be computed using the Stefan-Boltzmann law which states that the total radiant heat energy (J/s) emitted by a surface is proportional to the fourth power of its absolute temperature (K).
Mathematically, the law can be expressed as;E = σT^4where E is the total emitted energy, T is the absolute temperature in Kelvin, and σ is the
Stefan-Boltzmann constant
(5.67 × 10^−8 Wm^−2 K^−4).Thus, at temperatures of 5000 K, 5500 K, and 6000 K, the solar irradiance at the top of the earth's atmosphere can be calculated as follows;E_5000 = σT^4 = 5.67 × 10^−8 × (5000)^4 = 3.89 × 10^7 Wm^−2E_5500 = σT^4 = 5.67 × 10^−8 × (5500)^4 = 5.83 × 10^7 Wm^−2E_6000 = σT^4 = 5.67 × 10^−8 × (6000)^4 = 8.45 × 10^7 Wm^−2To compare the results obtained with those presented in Figures 2.9 and 2.10, the plots of the spectral solar irradiance as a function of wavelength for the three
temperatures
should be generated. The results can be compared based on the
wavelength
ranges and peak irradiance values obtained in the two figures.
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By comparing the computed values with the figures, we can analyze the differences and similarities in the solar irradiance at different temperatures.
To compute the solar irradiance at the top of the Earth's atmosphere emitted from temperatures of 5000, 5500, and 6000 K, we can use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its temperature.
The formula for the power radiated by a black body is given by [tex]\rm \(P = \sigma \cdot A \cdot T^4\)[/tex], where P is the power radiated, [tex]\(\sigma\)[/tex] is the Stefan-Boltzmann constant (approximately [tex]\rm \(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\)), \(A\)[/tex] is the surface area of the black body, and T is the temperature in Kelvin.
To compute the solar irradiance, we need to know the surface area of the Earth. Assuming the Earth to be a perfect sphere, its surface area can be calculated using the formula [tex]\(A = 4\pi R^2\)[/tex], where R is the radius of the Earth.
Substituting the values into the formula, we can calculate the solar irradiance for each temperature:
For [tex]\(5000 \, \text{K}\)[/tex]:
Solar irradiance [tex]\rm \(= \sigma \cdot A \cdot T^4\)[/tex]
Substituting the values, we get:
Solar irradiance [tex]\(= 5.67 \times 10^{-8} \cdot (4\pi R^2) \cdot (5000^4)\)[/tex]
Similarly, we can calculate the solar irradiance for temperatures of [tex]\(5500 \, \text{K}\) and \(6000 \, \text{K}\)[/tex].
To compare the results with Figures 2.9 and 2.10, we need to plot the computed solar irradiance values against the wavelength of the radiation. These figures show the solar irradiance spectrum at the top of the Earth's atmosphere for different wavelengths.
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What is the momentum of a two-particle system composed of a 1300 kg carmoving east at 40m / s and a second 900 kg car moving west at 85m / s ? Let east be the positive direction. Answer in units of kg m / s
The momentum of the two-particle system is -24500 kg m/s, opposite to the positive direction.
In a two-particle system, momentum is conserved. Here we have a 1300 kg car moving east at 40m/s and a second 900 kg car moving west at 85m/s. Let's find out the momentum of the system.
Mass of the 1st car, m1 = 1300 kg
Velocity of the 1st car, v1 = +40 m/s (east)
Mass of the 2nd car, m2 = 900 kg
Velocity of the 2nd car, v2 = -85 m/s (west)
Taking east as positive, the momentum of the 1st car is
p1 = m1v1 = 1300 × 40 = +52000 kg m/s
Taking east as positive, the momentum of the 2nd car is
p2 = m2v2 = 900 × (-85) = -76500 kg m/s
As the 2nd car is moving in the opposite direction, the momentum is negative.
The total momentum of the system,
p = p1 + p2 = 52000 - 76500= -24500 kg m/s
Therefore, the momentum of the two-particle system is -24500 kg m/s. The negative sign means the total momentum is in the west direction, opposite to the positive direction.
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You place an object 19 6 cm in front of a diverging lens which has a focal length with a magnitude of 13.0 cm. Determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 3.75. ______ cm
The object should be placed approximately 9.53 cm in front of the lens in order to produce an image that is reduced by a factor of 3.75.
To determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 3.75, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the lens
v is the image distance
u is the object distance
Given:
f = -13.0 cm (negative sign indicates a diverging lens)
v = -3.75u (image is reduced by a factor of 3.75)
Substituting these values into the lens formula, we have:
1/-13.0 = 1/(-3.75u) - 1/u
Simplifying the equation:
-1/13.0 = (1 - 3.75) / (-3.75u)
-1/13.0 = -2.75 / (-3.75u)
Cross-multiplying:
-1 * (-3.75u) = 2.75 * 13.0
3.75u = 35.75
Dividing by 3.75:
u ≈ 9.53 cm
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(a) Given a 36,0 V battery and 18.0 D and 92.0 resistors, find the current (in A) and power (in W) for each when connected in series. 19.00 P18.00 = A 192,00 P92.00 = W (b) Repeat when the resistances are in parallel 19.00 = P18.0 n = w TA 192.00 - P2.00 = w
(a) To find the current (in A) and power (in W) when connected in series,
we use the formula:
V = IRV = 36.0V
Resistor 1: R1 = 18.0Ω
Resistor 2: R2 = 92.0Ω
Equivalent resistance: RT = R1 + R2
= 18.0Ω + 92.0Ω
= 110.0ΩI
= V/R = 36.0V/110.0Ω
= 0.327 A19.00 P18.00 = A - The current is 0.327 A, which is the same through both resistors.
P = VI = (0.327 A)(36.0 V)
= 11.772 W - The power is 11.772 W for both resistors.
(b) When the resistances are in parallel, we use the formula:
1/RT = 1/R1 + 1/R21/RT
= 1/18.0Ω + 1/92.0Ω1/RT
= 0.062 + 0.011RC
= (1/0.062 + 0.011)-1
= 15.3ΩI1
= V/R1
= 36.0 V/18.0 Ω
= 2.0 AI2
= V/R2
= 36.0 V/92.0 Ω
= 0.391 A19.00 = P18.0
n = w - The current through the 18.0 Ω resistor is 2.0 A, and the current through the 92.0 Ω resistor is 0.391
A.T = P1 + P2 = V(I1 + I2) = (36.0 V)(2.0 A + 0.391 A) = 76.08 W - The total power is 76.08 W.
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A ball is thrown at an unknown angle. However a speed gon was able to deleet the ball's speed to be 30.0 m/s at the moment the ball was released from the persons hand. The release point is 1.89 m above the ground. If the ball lands a horizontal distance of 70 m away, what is the a) launch angle b) maximum height C) final velocity
Given information:Speed of the ball, v₀ = 30.0 m/sThe release point is 1.89 m above the ground.Horizontal distance, R = 70 m
a) Launch angleThe equation of motion of the ball can be represented as, R = v₀²sin2θ/g where g is the acceleration due to gravityR = 70 m, v₀ = 30 m/s, and g = 9.8 m/s²By substituting the given values, we get,70 = 30² sin2θ/9.8sin2θ = (70*9.8)/(30²)sin2θ = 0.4111θ = 0.4111/2 = 0.2057 radianUsing the radian to degree conversion formula,θ = 0.2057 * 180/π ≈ 11.8°Therefore, the launch angle is 11.8°.
b) Maximum heightThe maximum height attained by the ball can be calculated using the equation, h = v₀²sin²θ/2gBy substituting the given values, we get,h = 30²sin²(0.2057)/(2*9.8)h ≈ 9.08 mTherefore, the maximum height is 9.08 m.
c) Final velocityThe final velocity of the ball can be calculated using the formula, v = √(v₀² - 2gh)By substituting the given values, we get,v = √(30² - 2*9.8*1.89)v ≈ 26.5 m/sTherefore, the final velocity is 26.5 m/s.
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Index of refraction Light having a frequency in vacuum of 5.4×10 14
Hz enters a liquid of refractive index 2.0. In this liquid, its frequency will be:
When light with a frequency of 5.4×10^14 Hz enters a liquid with a refractive index of 2.0, its frequency will remain the same.
The frequency of light refers to the number of complete oscillations or cycles it undergoes per unit of time. The index of refraction, denoted by "n," is a property of a medium that describes how light propagates through it.
It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. In this case, the light enters a liquid with a refractive index of 2.0.
When light passes from one medium to another, its speed and wavelength change, while the frequency remains constant. The frequency of light is determined by the source and remains constant regardless of the medium it traverses.
Therefore, the frequency of light with a value of 5.4×10^14 Hz will remain the same when it enters the liquid with a refractive index of 2.0.In summary, the frequency of light with a vacuum frequency of 5.4×10^14 Hz will not change when it enters a liquid with a refractive index of 2.0.
The index of refraction only affects the speed and wavelength of light, while the frequency remains constant throughout different media.
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A2.3 kg wooden block is rest on a frictionless surface. A 25 g bullet traveling horizontally with a speed of 800 m/s penetrates and moves together with the wooden block. What is their velocity in m/s? 620 5.52 708 A stone is dropped from the top of a cliff. I is scen to hit the ground below after 9.3 seconds. Hong high is the cliff in meters? 415 433 424 442
The velocity of the block and bullet is 5.52 m/s.
Given data: Mass of the wooden block, m1 = 2.3 kgMass of the bullet, m2 = 25 g = 0.025 kg Velocity of the bullet, u = 800 m/sVelocity of the block and bullet, v = ?As the bullet penetrates the wooden block, the momentum of the system remains conserved before and after the collision.
Let u1 be the initial velocity of the block before the bullet hits it. Then, by conservation of momentum,m1u1 + m2u = (m1 + m2)v∴ v = (m1u1 + m2u) / (m1 + m2)Initially, the block is at rest. Therefore, u1 = 0. Substituting the values in the above equation, v = (0 + 0.025 x 800) / (2.3 + 0.025)≈ 5.52 m/s. Therefore, the velocity of the block and bullet after collision is 5.52 m/s. Hence, option 2 is correct. Let h be the height of the cliff. Given that the stone takes 9.3 seconds to hit the ground, the time of fall, t = 9.3 s.The stone falls freely under gravity, and the acceleration due to gravity, g = 9.8 m/s². Using the formula for the height of fall, we haveh = (1/2) × g × t²Hence,h = (1/2) × 9.8 × 9.3²≈ 415 m. Therefore, the height of the cliff is approximately 415 meters. Hence, option 1 is correct.
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a) A Hall-effect probe operates with a 107mA current. When the probe is placed in a uniform magnetic field with a magnitude of 0.0806T, it produces a Hall voltage of 0.689 μV. When it is measuring an unknown magnetic field, the Hall voltage is 0.352 μV. What is the unknown magnitude of the field?
b) If the thickness of the probe in the direction of B is 1.94mm, calculate the charge-carrier density (each of charge e).
(a) The unknown magnitude of the field is 0.00506 T.
(b) The charge-carrier density is 495 × 1019 m⁻³.
a) The Hall coefficient for the probe can be calculated using the equation: RH = VHB/I = 0.689μV/(107mA × 0.0806T) = 8.12×10⁻⁷ m³/C
The unknown magnetic field's magnitude can be determined using the equation: VB = RH × I × B0.352 × 10-6 V = 8.12 × 10⁻⁷ m³/C × 107 mA × BUnknown magnetic field, B = 0.00506 T
b) The charge-carrier density (n) can be calculated using the equation:n = 1/Re × e × μn, Where Re is the resistance of the material, e is the charge of an electron, and μn is the mobility of the material.
The resistance of the probe can be calculated using the equation: Re = l/(σt)where l is the length of the probe, t is the thickness of the probe in the direction of B, and σ is the conductivity of the material. Assuming the probe is rectangular in shape, we can use the equation: Re = w × h/(σt)where w is the width of the probe, and h is the height of the probe.
The area of the probe can be calculated using the equation:
A = w × h = t × w = 1.94 × 10⁻³ m²
The conductivity of the material can be calculated using the equation:σ = n × e2 × μ
The mobility of the material is given by the Hall coefficient equation:
RH = 1/ne = 1/Re × B
The charge-carrier density can now be calculated using the equation:n = 1/Re × e × μn = (B/Re × RH) × e × μn = (0.00506 T/Re × 8.12 × 10⁻⁷ m³/C) × 1.6 × 10⁻¹⁹ C × 0.001 m2/Vs = 495 × 1019 m⁻³
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A 1.95-kg particle has a velocity (1.96 1-3.03 j) m/s, and a 2.96-kg particle has a velocity (1.04 i +6.09 j) m/s. (a) Find the velocity of the center of mass. 1) m (b) Find the total momentum of the system. 1) kg- m/s + m/s
The velocity of the center of mass can be determined by dividing the total momentum of the system by the total mass.
The total momentum is calculated by summing the momentum (mass times velocity) of each particle.
To determine the velocity of the center of mass, we first calculate the momentum (product of mass and velocity) of each particle. Sum these momenta and divide by the total mass of the system. The total momentum of the system is the sum of the individual momenta.
Let's denote the masses and velocities as follows: m1 = 1.95 kg, v1 = (1.96 i - 3.03 j) m/s, m2 = 2.96 kg, v2 = (1.04 i + 6.09 j) m/s.
(a) The velocity of the center of mass Vcm is given by the formula: Vcm = (m1*v1 + m2*v2) / (m1 + m2).
(b) The total momentum P of the system is given by the sum of the momenta of each particle: P = m1*v1 + m2*v2.
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A 25 kg block is being pushed forward on a flat surface with a force of magnitude 66 N. The coefficient of static friction on the block is 0.23 and the coefficient of kinetic friction on the block is 0.16 (only one of these needs to be used). You are encouraged to draw a free body diagram of the block before trying the following questions. a) What is the net force acting on the block? b) What is the acceleration of the block?
The block has a coefficient of static friction of 0.23 and a coefficient of kinetic friction of 0.16. We must determine the net force acting on the block and its acceleration.
To solve this problem, we first draw a free-body diagram of the block. The forces acting on the block are the applied force pushing it forward, the gravitational force pulling it downward (mg), and the frictional force opposing its motion. The net force acting on the block is the vector sum of all the forces. In this case, the net force can be calculated as the applied force minus the force of friction. The force of friction can be determined by multiplying the coefficient of friction (either static or kinetic) by the normal force, which is equal to the weight of the block (mg). Therefore, the net force is given by
[tex]F_net = F_applied - μ * mg,[/tex]
where μ is the coefficient of friction.The acceleration of the block can be determined using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration [tex](F_net = ma)[/tex]
. Rearranging the equation,
we get [tex]a = F_net / m[/tex]
.By plugging in the given values into the equations, we can calculate the net force and the acceleration of the block.
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If A and B are vectors and B = -A, which of following is true? a) The magnitude of B is equal to the negative of the magnitude of Ā. b) A and B are perpendicular. c) The direction angle of B is equal to the direction angle of A plus 180°. d) A + B = 2 2. The lengths of vectors A, B, C and D are given by A = 75, B = 60,C = 25, and D = 90, and their direction angles are shown in the figure. a) Find the sum A + B + C + D in terms of its components. 30.07 27.01 52.0" b) What is the magnitude of the sum A + B + C + D? c) What is the direction of the sum A + B + C + Õ? (Angle from positive x-axis) 3. The measured length of a cylindrical laser beam is 12.1 meters, its measured diameter is 0.00121 meters, and its measured intensity is 1.21 x 10SW/m². Which of these measureme is the most precise? A. length B. diameter C. intensity D. All three are equally precise.
If B = -A, the correct statements are a) The magnitude of B is equal to the magnitude of A, but in the opposite direction, and c) The direction angle of B is equal to the direction angle of A plus 180°.
In the given scenario, A + B + C + D can be found by summing the respective components of the vectors. The magnitude of the sum can be calculated using the Pythagorean theorem, and the direction can be determined by finding the angle from the positive x-axis.
a) When B = -A, it means that the magnitude of B is equal to the magnitude of A, but in the opposite direction. Therefore, the statement "The magnitude of B is equal to the negative of the magnitude of Ā" is incorrect.
b) A and B being perpendicular is not necessarily true when B = -A. Perpendicular vectors have a dot product of zero, but in this case, the dot product of A and B would be negative, indicating an acute angle between them. Therefore, the statement "A and B are perpendicular" is incorrect.
c) When B = -A, the direction angle of B is equal to the direction angle of A plus 180°. This is because B is essentially the same vector as A but pointing in the opposite direction. Therefore, the statement "The direction angle of B is equal to the direction angle of A plus 180°" is correct.
In order to find the sum A + B + C + D in terms of its components, you would add the respective components of the vectors. Let's assume the components are given as (Ax, Ay), (Bx, By), (Cx, Cy), and (Dx, Dy). Then the sum of the components would be (Ax + Bx + Cx + Dx, Ay + By + Cy + Dy).
The magnitude of the sum A + B + C + D can be calculated using the Pythagorean theorem. If the components of the sum are (Sx, Sy), then the magnitude is given by √(Sx^2 + Sy^2).
The direction of the sum A + B + C + D can be determined by finding the angle from the positive x-axis. If the components of the sum are (Sx, Sy), the direction angle can be calculated using the arctan(Sy/Sx) formula. This will give the angle in radians.
To convert it to degrees, you can multiply by (180/π).
Regarding the last question about precision, the most precise measurement would be the one with the smallest relative uncertainty. Without the provided uncertainties or a better understanding of the measurement process, it is not possible to determine the most precise measurement among the given options (length, diameter, intensity). Therefore, the answer is D) All three measurements are equally precise until more information about the uncertainties is provided.
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the resistance of a 60cm wire of cross sectional area 6 x 10^-6m^2 is 200 ohms. what is the resistivity of the material of this wire
The resistivity of the material of the wire can be calculated using the formula: resistivity = (resistance x cross-sectional area) / length. In this case, the resistivity of the material is 3.33 x 10^-7 ohm-meter.
The resistivity of a material is a measure of how strongly it opposes the flow of electric current. It is denoted by the symbol ρ (rho). The resistivity can be calculated using the formula ρ = (R x A) / L, where R is the resistance, A is the cross-sectional area, and L is the length of the wire.
In this case, the given resistance is 200 ohms, the cross-sectional area is 6 x 10^-6 m^2, and the length of the wire is 60 cm (or 0.6 m). Plugging these values into the formula, we get ρ = (200 ohms x 6 x 10^-6 m^2) / 0.6 m = 2 x 10^-3 ohm-meter.
Therefore, the resistivity of the material of the wire is 3.33 x 10^-7 ohm-meter. The resistivity provides information about the intrinsic property of the material and can be used to compare the conductive properties of different materials.
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Which one of the following is NOT equal to the potential energy stored on a fully charged capacitor with a capacitance of C farads, connected to a battery of V volts, and holding Q coulombs of charge? X A. (Q.V) joules A volt is a joule per coulomb, so multiplying volts by coulombs yields joules. X B. 1 2 (2) joules C A volt is a joule per coulomb, and capacitance is expressed as coulombs per volt, so dividing the square of coulombs by farads yields joules. A volt is a joule per coulomb, and capacitance is expressed as coulombs per volt, so multiplying farads by the square of voltage yields joules. 1/2 ( 7² ) joules volt is a joule per coulomb, and capacitance is expressed as coulombs per volt, so dividing the square of farads by volts yields coulombs to the fifth power divided by joules to the third power, not joules. X C. (C.V²) joules O D. 1 Detailed Guidance Send us feedback. Feedback Info Capacitors behave according to the equation: C = Q V where C is capacitance in farads, Q is charge in coulombs, and Vis volts. Since a volt is defined as one joule per coulomb, the charge leaving a discharging capacitor has energy of Q. Vcap joules. The voltage of a fully charged capacitor is equal to the voltage of the battery that charged it, but when the capacitor is almost completely discharged its voltage is essentially zero. Because of this, the actual energy stored on a capacitor is equal to (1/2)(Q Vbatt). Each of the answer choices is equivalent to this value except (D), which, because it does NOT represent the energy stored by the capacitor, is correct.
The potential energy stored on a fully charged capacitor with capacitance C farads, connected to a battery of V volts, and holding Q coulombs of charge is not equal to the expression given in option D.
Capacitors store energy in the form of electric potential energy. The energy stored on a capacitor can be calculated using the equation E = (1/2)(QV), where E is the energy in joules, Q is the charge in coulombs, and V is the voltage in volts. The voltage across the capacitor is equal to the voltage of the battery that charged it.
In the given options, option D states that the energy stored on the capacitor is 1 joule. However, this is incorrect. The correct expression for the energy stored on the capacitor is (1/2)(QV), which is equivalent to option A, B, and C. Option D does not represent the energy stored by the capacitor.
Therefore, the correct answer is option D. The potential energy stored on a fully charged capacitor with a capacitance of C farads, connected to a battery of V volts, and holding Q coulombs of charge is not equal to 1 joule.
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What is the output voltage of a 3.00-V lithium cell in a digital wristwatch that draws 0.670 mA, if the cell's internal resistance is 2.25 Ω? (Enter your answer to at least five significant figures.) V
The output voltage of a 3.00-V lithium cell in a digital wristwatch, considering its internal resistance of 2.25 Ω, is approximately 1.5075 V which is determined using Ohm's Law and should be calculated to at least five significant figures.
To calculate the output voltage, we can use Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R). In this case, the current drawn by the wristwatch is given as 0.670 mA, and the internal resistance of the cell is 2.25 Ω. Thus, we can calculate the voltage as follows:
V = I * R
= 0.670 mA * 2.25 Ω
= 1.5075 mV
Since the given lithium cell has an initial voltage of 3.00 V, the output voltage will be slightly lower due to the internal resistance. Therefore, the output voltage of the lithium cell in the digital wristwatch is approximately 1.5075 V when rounded to five significant figures.
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A single-phase 40-kVA, 2000/500-volt, 60-Hz distribution transformer is used as a stepdown transformer. Winding resistances are R1 = 2 Ω and R2 = 0.125 Ω; leakage reactances are X1 = 8 Ω and X2 = 0.5 Ω. The load resistance on the secondary is 12 Ω. The applied voltage at the terminals of the primary is 1000 V. (a) Replace all circuit elements with perunit values. (b) Find the per-unit voltage and the actual voltage V2 at the load terminals of the transformer
The problem involves a single-phase distribution transformer with specified ratings and parameters. The task is to convert the circuit elements to per-unit values and calculate the per-unit voltage and the actual voltage at the load terminals of the transformer.
In the given problem, a single-phase 40-kVA, 2000/500-volt, 60-Hz distribution transformer is considered. The transformer is used as a step-down transformer, and its winding resistances and leakage reactances are provided. The load resistance on the secondary side is given as 12 Ω, and the applied voltage at the primary terminals is 1000 V.
To analyze the transformer on a per-unit basis, all circuit elements need to be converted to per-unit values. This involves dividing the actual values by the base values. The base values are typically chosen as the rated values of the transformer. In this case, the base values can be taken as 40 kVA, 2000 volts, and 12 Ω.
By dividing the actual values of resistances and reactances by their corresponding base values, the per-unit values can be obtained. Similarly, the load resistance on the secondary side can be expressed per per-unit by dividing it by the base resistance. After converting the circuit elements to per-unit values, the per-unit voltage can be calculated by dividing the applied voltage at the primary terminals by the base voltage. This provides a relative value that can be used for further calculations.
To find the actual voltage at the load terminals of the transformer, the per-unit voltage is multiplied by the base voltage. This gives the actual voltage value in volts. In conclusion, the problem involves converting the circuit elements of a distribution transformer to per-unit values and calculating the per-unit voltage and the actual voltage at the load terminals. This analysis allows for a standardized representation of the transformer's parameters and facilitates further calculations and comparisons.
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what is the rate of motion longitudal AND lateral in mm per year
and direction of the plates moving
GPS Time Series Database. The JPL website references the Cocos Plate as ISCO in their database. If you'd like to see the actual cell-tower, use the blue-numbers below: paste the coordinates into Googl
The rate of motion longitudal and lateral in mm per year and direction of the plates moving are essential concepts in plate tectonics. Plate tectonics is a geologic theory that explains the Earth's crust and its movements.
There are a variety of directions in which tectonic plates are moving. The Pacific plate, for example, is moving in a westerly direction. It's worth noting that while tectonic plates are always in motion, their motion is not always constant. The longitudinal and lateral movements of tectonic plates occur at varying rates. The rate of motion is typically expressed in millimeters per year. The speed of the plates' motion, as well as their direction, may vary depending on the location of the tectonic plates and the forces acting on them. Tectonic plates are either converging, diverging, or slipping against one another at their boundaries. The type of plate boundary, whether convergent, divergent, or transform, determines the rate and direction of plate motion.
Longitudinal motion or movement is defined as the movement of plates in a direction parallel to the boundary or toward or away from each other. The Pacific Plate is currently moving in a northwest direction at a rate of about 100 mm per year. Lateral motion or movement, on the other hand, is the movement of plates in a direction perpendicular to the boundary. The boundary between the North American Plate and the Pacific Plate, for example, runs roughly parallel to the Pacific Northwest coastline and is slipping sideways or moving horizontally at a rate of about 40 mm per year. Therefore, the rate of motion longitudal and lateral in mm per year is dependent on the location of the tectonic plates and the forces acting on them.
Tectonic plates are in constant motion, moving longitudinally and laterally at varying rates and directions depending on their location and the type of boundary.
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DETAILS SERCP10 27.P.009. 0/4 Submissions Used MY NOTES ASK YOUR TEACHER When light of wavelength 140 nm falls on a carbon surface, electrons having a maximum kinetic energy of 3.87 eV are emitted. Find values for the following. (a) the work function of carbon ev (b) the cutoff wavelength nm (c) the frequency corresponding to the cutoff wavelength Hz Additional Materials eBook
The photoelectric effect demonstrates the particle-like properties of light, where photons interact with electrons on a surface.
The work function of carbon, cutoff wavelength, and frequency corresponding to the cutoff wavelength can be determined using this principle, given the incoming light's wavelength and the maximum kinetic energy of emitted electrons. For a more detailed explanation, the energy of a photon is given by the formula E=hf, where h is Planck's constant and f is the frequency of light. The energy of a photon can also be expressed as E=(hc/λ), where λ is the wavelength. The work function (φ) is the minimum energy required to remove an electron from the surface of a material. According to the photoelectric effect, the energy of the incoming photon is used to overcome the work function, and the rest is given to the electron as kinetic energy. Thus, hc/λ - φ = KE. Substituting given values, we can solve for φ. For cutoff wavelength, we consider when KE=0, implying φ=hc/λ_cutoff. Rearranging and substituting φ, we can find λ_cutoff. The frequency corresponding to the cutoff wavelength is simply c/λ_cutoff.
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A swimmer is swimming at 1 knot (nautical miles per hour) on a heading of N30⁰W. The current is
flowing at 2 knots towards a bearing of N10⁰E. Find the velocity of the swimmer, relative to the shore.
The magnitude of the swimmer's velocity relative to the shore is approximately 1.199 knots, and the direction is approximately N86.18⁰W. To find the velocity of the swimmer relative to the shore, we can break down the velocities into their components and then add them up.
Swimmer's velocity: 1 knot on a heading of N30⁰W
Current's velocity: 2 knots towards a bearing of N10⁰E
First, let's convert the velocities from knots to a common unit, such as miles per hour (mph). 1 knot is approximately equal to 1.15078 mph.
Swimmer's velocity:
1 knot = 1.15078 mph
Current's velocity:
2 knots = 2.30156 mph
Swimmer's velocity:
[tex]Velocity_N[/tex] = 1 knot * cos(30⁰) = 1 knot * √(3)/2 ≈ 0.866 knots
[tex]Velocity_W[/tex] = 1 knot * sin(30⁰) = 1 knot * 1/2 ≈ 0.5 knots
Current's velocity:
[tex]Velocity_N[/tex] = 2 knots * sin(10⁰) = 2 knots * 1/6 ≈ 0.333 knots
[tex]Velocity_E[/tex] = 2 knots * cos(10⁰) = 2 knots * √(3)/6 ≈ 0.577 knots
Now, we can add up the north-south and east-west components separately to find the resultant velocity relative to the shore.
Resultant [tex]velocity_N[/tex] = [tex]velocity_N[/tex] (swimmer) + [tex]velocity_N[/tex] (current) ≈ 0.866 knots + 0.333 knots ≈ 1.199 knots
Resultant [tex]velocity_W[/tex] = [tex]velocity_W[/tex] (swimmer) - [tex]Velocity_E[/tex] (current) ≈ 0.5 knots - 0.577 knots ≈ -0.077 knots
Note that the negative value indicates that the resultant velocity is in the opposite direction of the west.
Finally, we can calculate the magnitude and direction of the resultant velocity using the Pythagorean theorem and trigonometry.
Resultant velocity = √(Resultant [tex]velocity_N^2[/tex]+ Resultant [tex]velocity_W^2[/tex])
≈ √((1.199 [tex]knots)^2[/tex]+ (-0.077 [tex]knots)^2[/tex]) ≈ √(1.437601 [tex]knots)^2[/tex] ≈ 1.199 knots
The direction of the resultant velocity relative to the shore can be determined using the arctan function:
Resultant direction = arctan(Resultant [tex]velocity_N[/tex]/ Resultant [tex]velocity_W[/tex])
≈ arctan(1.199 knots / -0.077 knots) ≈ -86.18⁰
Therefore, the magnitude of the swimmer's velocity relative to the shore is approximately 1.199 knots, and the direction is approximately N86.18⁰W.
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What is the magnification for a simple magnifier of focal length 5 cm, assuming the user has a normal near point of 25 cm ? 5 25 12.5 125
the magnification for a simple magnifier of focal length 5 cm, assuming the user has a normal near point of 25 cm is 6.Please note that the answer is 75 words.
The magnification for a simple magnifier of focal length 5 cm, assuming the user has a normal near point of 25 cm is 5. This can be computed using the formula:
Magnification of simple microscope = (D/f) + 1, where D is the least distance of clear vision or near point, and f is the focal length of the lens or magnifying glass.
Given that focal length of simple magnifier, f = 5 cmLeast distance of clear vision, D = 25 cmMagnification = (25/5) + 1= 5 + 1= 6
Therefore, the magnification for a simple magnifier of focal length 5 cm, assuming the user has a normal near point of 25 cm is 6.Please note that the answer is 75 words.
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.1. It takes you 10 min to walk with an average velocity of 2 m/s to The North from The Grocery Shop to your house. What is your displacement? 2. Two buses, A and B, are traveling in the same direction, although bus A is 200 m behind bus B. The speed of A is 25 m/s, and the speed of B is 20 m/s. How much time does it take for A to catch B ? 3. A truck accelerates from 10 m/s to 20 m/s in 5sec. What is it acceleration? How far did it travel in this time? Assume constant acceleration. 4. With an average acceleration of −2 m/s^2
, how long will it take to a cyclist to bring a bicycle with an initial speed of 5 m/s to a complete stop? 5. A car with an initial speed of 5 m/s accelerates at a uniform rate of 2 m/s ^2
for 4sec. Find the final speed and the displacement of the car during this time. 6. You toss a ball straight up with an initial speed of 40 m/s. How high does it go, and how long is it in the air (neglect air drag)?
1. To find the displacement, we use the formula:
Displacement = Velocity × Time
= 2 m/s × 10 min × 60 s/min
= 1200 m
Therefore, the displacement is 1200 m to the North.
2. The distance that A has to cover to catch up with B is 200 m. Let t be the time it takes for A to catch up with B. Then the distance each bus covers will be:
Distance covered by bus A = Speed of bus A × Time = 25 m/s × t.
Distance covered by bus B = Speed of bus B × Time + Distance between them = 20 m/s × t + 200 m.
As the buses are moving in the same direction, A will catch up with B when the distance covered by A is equal to the distance covered by B. Therefore, we can set these two equations equal to each other:
25t = 20t + 200.
This simplifies to 5t = 200, which gives us t = 40 seconds.
Therefore, it will take A 40 seconds to catch up with B.
3. To find the acceleration, we use the formula:
Acceleration = (Final Velocity − Initial Velocity) ÷ Time
= (20 m/s − 10 m/s) ÷ 5 s
= 2 m/s^2.
To find the distance, we use the formula:
Distance = (Initial Velocity × Time) + (0.5 × Acceleration × Time^2)
= (10 m/s × 5 s) + (0.5 × 2 m/s^2 × (5 s)^2)
= 25 m + 25 m
= 50 m.
Therefore, the acceleration is 2 m/s^2 and the distance traveled is 50 m.
4. To find the time taken to stop, we use the formula:
Final Velocity = Initial Velocity + (Acceleration × Time).
As the final velocity is 0 (since the cyclist is coming to a complete stop), we can rearrange this formula to solve for time:
Time = (Final Velocity − Initial Velocity) ÷ Acceleration
= (0 − 5 m/s) ÷ −2 m/s^2
= 2.5 seconds.
Therefore, it will take 2.5 seconds for the cyclist to bring the bicycle to a complete stop.
5. To find the final speed, we use the formula:
Final Velocity = Initial Velocity + (Acceleration × Time)
= 5 m/s + (2 m/s^2 × 4 s)
= 13 m/s.
To find the displacement, we use the formula:
Displacement = (Initial Velocity × Time) + (0.5 × Acceleration × Time^2)
= (5 m/s × 4 s) + (0.5 × 2 m/s^2 × (4 s)^2)
= 20 m + 16 m
= 36 m.
Therefore, the final speed is 13 m/s and the displacement is 36 m.
6. When the ball is at its maximum height, its final velocity is 0 m/s. Therefore, we can use the formula:
Final Velocity = Initial Velocity + (Acceleration × Time).
As the final velocity is 0 and the initial velocity is 40 m/s, we can solve for time:
Time = Final Velocity ÷ Acceleration
= 40 m/s
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Give your answers in SI units and to three significant figures. Question 1 3 pts Newer automobiles have filters that remove fine particles from exhaust gases. This is done by charging the particles and separating them with a strong electric field. Consider a positively charged particle +8μC that enters an electric field with strength 6×10 6
N/C. The particle is traveling at 77 m/s and has a mass of 1 g. If the horizontal width of the filter is 20 cm, determine the vertical distance that the particle will be deflected as it passes through the filter. Express your answer in meters.
The vertical distance that the particle will be deflected as it passes through the filter is 7.09 x 10^-6 m.
Explanation:Given,Charge of the particle, q = +8μC = +8 × 10^-6 CStrength of electric field, E = 6 × 10^6 N/CVelocity of the particle, v = 77 m/sMass of the particle, m = 1 g = 10^-3 kgWidth of the filter, d = 20 cm = 0.2 mThe electric force acting on a charged particle in an electric field is given byF = qE ……… (1)The particle will experience force in the horizontal direction, F = qE ……… (2)It will move with constant velocity in the vertical direction and experiences force of gravity in the vertical direction, F = mg ……… (3)Let ‘y’ be the vertical deflection. Net force experienced by the particle along the y-axis is given asFy = mg ……… (4)By Newton’s second law, F = ma ……… (5)Net force experienced by the particle along the x-axis is given asFx = qE ……… (6)Net force acting on the particle is given asFnet = √(Fx^2 + Fy^2) ……… (7)The net force acting on the particle is given asqE = ma ……… (8).
As the particle is moving with constant velocity along the y-axis, its acceleration along the y-axis is zero.Therefore, Fy = 0mg = 0y = 0Also, the net force acting on the particle is given by, Fnet = qE ……… (9)Fnet = qE = +8 × 10^-6 × 6 × 10^6 = 48 × 10^-6 NNet force acting on the particle along the x-axis is given as,Fx = Fnet sin θ ……… (10)θ = tan^-1 (y/d)Fx = ma = Fnet cos θ ……… (11)θ = tan^-1 (y/d)a = Fnet/m = (qE)/mcos θsin θ = y/dcos θ = √(1 – sin^2 θ)cos θ = √(1 – (y/d)^2)Fx = ma = Fnet cos θ(8 × 10^-6) × 6 × 10^6 √(1 – (y/0.2)^2) = (10^-3) × ay/0.2 = (48 × 10^-6)/[(10^-3) × 6 × 10^6 √(1 – (y/0.2)^2)]y = 7.09 x 10^-6 m.
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Two railroad cars are about to collide. One is stationary (v=0) and has a mass of 5000 kg.
The other one is moving left towards it 2 m/s and its mass is 2000 kg. Assuming it is a
totally inelastic collision, how fast and what direction will the two cars be moving after the
collision?
After the collision, the two railroad cars will move together at a final velocity of 4/7 m/s in the leftward direction.
In the given scenario, two railroad cars, one stationary and one moving leftwards at 2m/s, with masses of 5000 kg and 2000 kg respectively, are about to collide.
Since the collision is inelastic, the two objects will stick together and move together after the collision at a common speed.
Let the final common speed of both objects be v. Applying the principle of conservation of momentum, we have:
Initial momentum = Final momentum (5000 kg) × (0 m/s) + (2000 kg) × (−2 m/s) = (5000 kg + 2000 kg) × v
∴ −4000 = 7000v
v = −4000 / 7000 = −4/7 m/s
As the final velocity is negative, this indicates that the combined object will move to the left, which is the direction of the initial velocity of one of the objects.
Hence, the final velocity of the combined object is 4/7 m/s leftwards.
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An ultra-fast pulse lasers emits pulses of 13 fs. The length of each pulse train is: A) 7.79 pm B) 3.9 pm C) 19.49 pm D 11.69 pm ) E) 3.9 pm Air
An ultra-fast pulse lasers emits pulses of 13 fs. The length of each pulse train is: The correct answer would be that there is not enough information given to determine the length of each pulse train (option O).
To determine the length of each pulse train emitted by the ultra-fast pulse laser, we need to consider the relationship between the pulse duration and the pulse repetition rate.
The length of each pulse train is given by the formula:
Length of each pulse train = Pulse duration × Pulse repetition rate
The pulse duration is provided as 13 fs (femtoseconds). However, the pulse repetition rate is not given in the question. Without knowing the pulse repetition rate, we cannot accurately determine the length of each pulse train.
Therefore, based on the information provided, we cannot determine the exact length of each pulse train emitted by the ultra-fast pulse laser. The correct answer would be that there is not enough information given to determine the length of each pulse train (option O).
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what would happen if a permanent magnet is placed on top of a straight wire
When a permanent magnet is placed on top of a straight wire, a magnetic field is produced in the region surrounding the wire due to the motion of charges in the wire. The magnetic field produced by the wire interacts with the magnetic field of the permanent magnet and causes a force to be exerted on the wire.
The direction of the force is perpendicular to both the magnetic field and the current in the wire. If the wire is not fixed in place, it will experience a force and move in a direction that is perpendicular to both the magnetic field and the current in the wire. This phenomenon is known as the Lorentz force, which is the force that is exerted on a charged particle when it moves in an electromagnetic field.
The direction of the force is given by the right-hand rule, which states that if the thumb of the right hand points in the direction of the current, and the fingers point in the direction of the magnetic field, then the palm of the hand will point in the direction of the force. The magnitude of the force is proportional to the current in the wire and the strength of the magnetic field.
Therefore, the stronger the magnetic field or the current, the greater the force that is exerted on the wire. The Lorentz force is the basis for the operation of many devices such as motors, generators, and transformers.
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A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F=42.0 N that is directed at an angle of 43.0 ∘
below the horizontal and the chair slides along the floor. Use Newton's laws to calculate the normal force that the floor exerts on the chair
The normal force that the floor exerts on the chair is approximately 107.9 N.
Mass of chair, m = 15 kgForce, F = 42.0 NAngle, θ = 43°Normal force, N is given by,Newton’s second law of motion states that the force acting on an object is directly proportional to the acceleration produced in it and inversely proportional to its mass.
It is given by, `F = ma`Where, F is the net force applied on the object, m is the mass of the object and a is the acceleration produced in the object. When an object is in contact with a surface, it experiences two types of forces:Normal force (N)Frictional force (f)According to Newton’s third law of motion, the normal force acting on an object is equal in magnitude and opposite in direction to the force applied by the object on the surface in contact.
Hence,Normal force, N = Force applied by the object on the surface in contactLet N1 be the normal force acting on the chair. From the free-body diagram of the chair, we can write,N1 + Fsinθ = mgwhere, m is the mass of the chair, g is the acceleration due to gravity and Fsinθ is the component of force F acting parallel to the surface.
Substituting the given values in the above equation, we getN1 = mg - Fsinθ= (15 kg) × (9.8 m/s²) - (42 N) × sin 43°≈ 107.9 N.
Therefore, the normal force that the floor exerts on the chair is approximately 107.9 N.
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A machine is placed on member BC which has an unbalanced force of 500 kg which varies sinusoidally. Neglecting the mass of the machine, determine: (i) the maximum displacement when the unit"s speed is 150rpm; (ii) the speed of the machine at resonance; (iii) the displacement at resonance. Note: Take the following values: - EI=20×10 3
kNm 2
- M=20 tonnes: - Consider BC as infinitely rigid.
Hence, the maximum displacement is 10.57 m, the speed of the machine at resonance is 2.5 rad/s, and the displacement at resonance is approximately 7.5 m.
The equation of motion is given as below: EI(d2y/dx2) = (Mx - 500cos ωt)yLet's integrate both sides, we get EI(dy/dx) = (Mx2/2 - 500cos ωt y2/2)dxWe know EI(d2y/dx2) = (d/dx)[EI(dy/dx)] and also d/dx(x2y2) = y2 + 2xy(dy/dx) + x2(d2y/dx2)So, on integrating,
we get EI(dy/dx) = (Mx2/2 - 500cos ωt y2/2)dx is equal to EI(dy/dx) = (M/3 x3 - 500/ωcos ωt y2)x + C1where C1 is a constant of integration.Let the maximum displacement occurs at x = x1when the unit's speed is 150 rpm.
Therefore, the equation of motion can be written as EI(d2y/dx2) = (Mx1 - 500)ySo, the maximum displacement is given by ym = Mx1/500Since the speed of the machine at resonance is given by ωn = [√(M/ EI)]/2π, the speed of the machine is given by ωn = [√(20000/ 20 × 106)]/2π = 2.5 rad/sAt resonance, EI(d2y/dx2) = My, so EI(d2y/dt2) = -Mωn2y = -500y
Thus, the displacement at resonance is given by y = ym/√(1 - (f/ fn)2)where fn = (ωn/2π) = 0.398 Hzf = 150 rpm = 2.5 Hz Therefore, f/fn = 6.29 so that y = ym/√(1 - (6.29)2) = 0.707ym = 10.57 m, at resonance, the displacement is given by y = 0.707 × 10.57 = 7.47m, approximately 7.5 m.
Hence, the maximum displacement is 10.57 m, the speed of the machine at resonance is 2.5 rad/s, and the displacement at resonance is approximately 7.5 m.
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Answer Both Parts Or Do Not Answer:
A uniform electric field is directed in the +x-direction and has a magnitude E. A mass 0.072 kg and charge +2.90 mC is suspended by a thread between the plates. The tension in the thread is 0.84 N.
What angle does the thread make with the vertical axis? Please give answer in degrees.
Find the magnitude of the electric force. Give answers in N to three significant figures.
The angle between the thread and the vertical axis is approximately 41.7 degrees. The magnitude of the electric force depends on the value of the electric field (E) and cannot be determined without that information.
To determine the angle the thread makes with the vertical axis, we can use trigonometry. The tension in the thread provides the vertical component of the force, and the electric force provides the horizontal component.
Given:
Mass (m) = 0.072 kg
Charge (q) = 2.90 mC = 2.90 × 10^(-3) C
Tension in the thread (T) = 0.84 N
The vertical component of the force is equal to the tension in the thread, so we have:
Tension (T) = mg
Solving for g, the acceleration due to gravity:
g = T / m
Substituting the values:
g = 0.84 N / 0.072 kg = 11.67 N/Kg
Next, we can find the magnitude of the electric force (F_e) using the formula:
F_e = qE
Given that the electric field magnitude (E) is directed in the +x-direction and has a value E, we can substitute the values:
F_e = (2.90 × 10^(-3) C) × E
The angle between the tension and the vertical axis can be found using the tangent function:
tan(theta) = Tension_y / Tension_x
tan(theta) = Weight / Tension
tan(theta) = 0.7056 N / 0.84 N
theta ≈ 41.7 degrees
Now, we can solve for θ by taking the inverse tangent (arctan) of both sides.
The magnitude of the electric force is given by F_e = (2.90 × 10^(-3) C) × E, where E is the electric field magnitude.
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For f = (2y-z)³ i + x² j - (3x²+1)k, is f conservative
at point (1,4,6)?
is there a curl?
is there a divergence?
For f = (2y-z)³ i + x² j - (3x²+1)k, is f conservative
at point (1,4,6)?
Curl (or rotation) is the curl of a vector field, which describes the magnitude and direction of the rotation of a particle at a point. To find whether f is conservative, we must find the curl of f and check whether it is zero or not.
The curl of the given function is: curl(f) = (∂Q/∂y - ∂P/∂z) i + (∂R/∂z - ∂P/∂x) j + (∂P/∂y - ∂Q/∂x) k
Where, P = (2y - z)³Q = x²R = -(3x² + 1)∂P/∂x = 0∂P/∂y = 6(2y - z)²∂P/∂z = -3(2y - z)²∂Q/∂x = 2x∂Q/∂y = 0∂Q/∂z = 0∂R/∂x = -6x∂R/∂y = 0∂R/∂z = 0
Therefore, curl(f) = (12z - 24y) i + 0 j + 6x k
At point (1, 4, 6),curl(f) = (12(6) - 24(4)) i + 0 j + 6(1) k= -72 i + 6 k
Therefore, the curl of f at point (1, 4, 6) is not zero. Therefore, f is not conservative at point (1, 4, 6).
Divergence is the measure of the magnitude of a vector field's source or sink at a given point in the field. To determine if there is a divergence, we must take the divergence of the function.
The divergence of the given function is:div(f) = ∂P/∂x + ∂Q/∂y + ∂R/∂z= 0 + 0 - 6
Therefore, the divergence of f is -6.
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Where are the young stars in spiral galaxies? In the disk. In the bulge. In the halo. Question 24 Where are the young stars in elliptical galaxies? In the bulge. In the disk. There are none. Question 25 Where are stars formed in our galaxy? In the halo. In the disk In the bulge
23. Young stars in spiral galaxies are typically found in the disk.
24. in the elliptical galaxies a few new stars might show up in the bulge
25. Stars are formed in the disk of our galaxy.
What should you know about the Elliptical galaxies?Elliptical galaxies are generally composed of older stars, with little to no ongoing star formation. This is due to the fact that they have used up or lost most of their interstellar medium. So, there are typically no young stars in elliptical galaxies.
Our galaxy, the Milky Way, is a barred spiral galaxy.
Stars are primarily formed in the disk of our galaxy, particularly in the spiral arms where the interstellar medium is densest.
This is where new stars, including young blue stars and star clusters, are most frequently born. The bulge and halo regions of the Milky Way are primarily composed of older stars, with very little ongoing star formation.
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