(a) Mass of the airplane, Therefore, the mass of the airplane is 1.47 × 10⁵ kg. (b)Force exerted by the tractor on the airplane. Therefore, the force exerted by the tractor on the airplane is 2.59 × 10⁴ N.
(a)Mass of the airplane the free-body diagram (FBD) is shown below:
The sum of the forces in the horizontal direction is given by:
ΣFx = maxFtrac - Ff = max
Rearranging the above equation in terms of the mass of the airplane, m, gives:m = (Ftrac - Ff) / a
Substituting the given values, Ftrac = 2.38 × 10⁴ N, Ff = 2400 N, and a = 0.150 m/s²m = (2.38 × 10⁴ - 2400) / 0.150m = 1.47 × 10⁵ kg
Therefore, the mass of the airplane is 1.47 × 10⁵ kg.
(b)Force exerted by the tractor on the airplane
The free-body diagram (FBD) is shown below:The sum of the forces in the horizontal direction is given by:
ΣFx = maxFtrac - Ff - Fplane = max
where Fplane is the force exerted by the airplane on the tractor. Since the airplane is being pushed backwards by the tractor, the force exerted by the airplane on the tractor is in the forward direction.
Substituting the given values,Ftrac = 2.38 × 10⁴ N, Ff = 2400 N, a = 0.150 m/s², and Ff(plane) = 2200 N,m = 1.47 × 10⁵ kg
Thus,2.38 × 10⁴ - 2400 - 2200 = (1.47 × 10⁵) × 0.150 × FplaneFplane = 2.59 × 10⁴ N
Therefore, the force exerted by the tractor on the airplane is 2.59 × 10⁴ N.
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A car with a mass of 750 kg moving at a speed of 23 m/s rear-ends a truck with a mass of 1250 kg and a speed of 15 m/s. (The two vehicles are initially traveling in the same direction.) If the collision is elastic, find the final velocities of the two vehicles. (This is a 1-dimensional collision.)
The final velocities of the two vehicles, if the collision is elastic, then v₁ = 18 m/s and v₂ = 48 m/s.
It is given that, Mass of car, m₁ = 750 kg, Initial velocity of car, u₁ = 23 m/s, Mass of truck, m₂ = 1250 kg, Initial velocity of truck, u₂ = 15 m/s and the collision is elastic. Therefore, the total momentum of the system is conserved, i.e.,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Putting the values, we get,
750 × 23 + 1250 × 15 = 750v₁ + 1250v₂
(17250 + 18750) = (750v₁ + 1250v₂)
36000 = 750v₁ + 1250v₂
(6 × 6000) = 750v₁+ 1250v₂
Now, we have two variables and only one equation. We need another equation. We can use the conservation of kinetic energy to get another equation.
Since the collision is elastic, the total kinetic energy of the system is conserved, i.e.,
(1/2)m₁*2u₁ + (1/2)m₂*2u₂ = (1/2)m₁*2v₁ + (1/2)m₂*2v₂
Putting the values, we get,
(1/2) × 750 × (23)2 + (1/2) × 1250 × (15)2 = (1/2) × 750 × 2v₁ + (1/2) × 1250 × 2v₂
Solving further, we get,
195375 = 375v₁ + 937.5v₂(195375 / 375) = v₁ + (937.5 / 375)v₂(521 / 5) = v₁ + (25 / 2)v₂
Multiplying the first equation by 25 and subtracting the second equation, we get,
15000 = (625/2)v₂
v₂ = 48 m/s
Putting the value of v₂ in the first equation, we get,
6 × 6000 = 750v1 + 1250(48)
v₁ = 18 m/s
Therefore, the final velocities of the two vehicles are:v₁ = 18 m/s , v₂= 48 m/s.
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Just then, you realized something---the wavelength of this man's butt beam is 525 nm. Didn't your pendulum have the print "project 525?" Was that a coincidence? When you confronted him, he said "I was just funding project 525. I was told to produce and sell as many free electrons as possible. Muons and antimuons have mean life (not half life) of 2.20 us, so it didn't take me a long time to produce 600 electrons from 1000 muons/antimuons that I was given." How long did it actually take him to do that? O 1.62 us 02.91 us 01.12 us 2.02 us
It took the man 880 μs (or 0.88 μs) to produce 600 electrons from the given 1000 muons/antimuons.
The man funded Project 525, which involved producing and selling free electrons. He was given 1000 muons/antimuons, and he managed to produce 600 electrons. Since muons and antimuons have a mean life (not half-life) of 2.20 μs, we can calculate the time it took for him to produce 600 electrons.
The mean life (τ) of a particle is related to its decay rate (λ) by the equation τ = 1/λ. In this case, the mean life of muons/antimuons is given as 2.20 μs.
The decay rate can be calculated using the formula λ = N/t, where N is the number of decays and t is the time interval. In this case, the number of decays is 1000 - 600 = 400, as 600 electrons were produced from the given 1000 muons/antimuons.
We can rearrange the formula to find the time interval: t = N/λ. Substituting the values, we have t = 400 / (1/2.20 μs) = 400 * (2.20 μs) = 880 μs.
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Electrical Principles [15] 2.1 An electric desk furnace is required to heat 0,54 kg of copper from 23,3°C to a melting point of 1085°C and then convert all the solid copper into the liquid state (melted state). The whole process takes 2 minutes and 37 seconds. The supply voltage is 220V and the efficiency is 67,5%. Assume the specific heat capacity of copper to be 389 J/kg.K and the latent heat of fusion of copper to be 206 kJ/kg. The cost of Energy is 236c/kWh. 2.1.1 Calculate the energy consumed to raise the temperature and melt of all of the copper.
The energy consumed to raise the temperature and melt all of the copper is 337196.182 J or 0.0937 kWh, and the total cost of energy consumed is 0.0221 R.
The electrical energy consumed to raise the temperature and melt all of the copper is calculated as follows:
Initial temperature of copper, T[tex]_{1}[/tex]= 23.3°C
Final temperature of copper, T[tex]_{2}[/tex] = 1085°C
Specific heat capacity of copper, c = 389 J/kg.K
Latent heat of fusion of copper, L[tex]_{f}[/tex] = 206 kJ/kg
Mass of copper, m = 0.54 kg
Time taken, t = 2 minutes 37 seconds = 157 seconds
Efficiency, η = 67.5% = 0.675
Supply voltage, V = 220 V
Cost of energy, CE = 236 c/kWh = 0.236 R/kWh
The energy required to raise the temperature of the copper from T[tex]_{1}[/tex] to T[tex]_{2}[/tex] is given by:
Q[tex]_{1}[/tex] = mc(T[tex]_{2}[/tex] - T[tex]_{1}[/tex])= 0.54 × 389 × (1085 - 23.3) = 0.54 × 389 × 1061.7= 225956.182 J
The energy required to melt the copper is given by:
Q[tex]_{2}[/tex] = mL[tex]_{f}[/tex]= 0.54 × 206 × 1000Q[tex]_{2}[/tex] = 111240 J
The total energy consumed is the sum of Q[tex]_{1}[/tex] and Q[tex]_{2}[/tex], that is:
Q[tex]_{tot}[/tex] = Q[tex]_{1}[/tex] + Q[tex]_{2}[/tex] = 225956.182 + 111240= 337196.182 J
The energy consumed is then converted from Joules to kWh:
Energy (kWh) = Q[tex]_{tot}[/tex] ÷ 3.6 × 10⁶
Energy (kWh) = 337196.182 ÷ 3.6 × 10⁶
Energy (kWh) = 0.0937 kWh
The total cost of energy consumed is calculated by multiplying the energy consumed (in kWh) by the cost of energy (in R/kWh):
Cost = Energy × CE = 0.0937 × 0.236
Cost = 0.0221 R
Therefore, the energy consumed to raise the temperature and melt all of the copper is 337196.182 J or 0.0937 kWh, and the total cost of energy consumed is 0.0221 R.
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8. [-12 Points] DETAILS SERCP11 22.7.P.037. A plastic light pipe has an index of refraction of 1.66. For total internal reflection, what is the mi (a) air 0 (b) water O Need Help? Read It MY NOTES ASK YOUR TEACHER internal reflection, what is the minimum angle of incidence if the pipe is in the following media? V MY NOTES ASK YOUR TEACHER
A plastic light pipe has an index of refraction of 1.66. for both (a) air and (b) water as the initial medium, total internal reflection does not occur when light enters the plastic light pipe with a refractive index of 1.66.
To determine the critical angle for total internal reflection, we can use Snell's law, which relates the angles of incidence and refraction at the interface between two media:
n1 × sin(theta1) = n2 × sin(theta2)
where:
n1 is the refractive index of the first medium (initial medium),
theta1 is the angle of incidence,
n2 is the refractive index of the second medium (final medium), and
theta2 is the angle of refraction.
For total internal reflection, the angle of refraction (theta2) becomes 90 degrees. Therefore, we can rewrite Snell's law as:
n1 × sin(theta1) = n2 × sin(90)
Since sin(90) = 1, the equation simplifies to:
n1 × sin(theta1) = n2
(a) Air as the initial medium:
Given n1 = 1 (approximating the refractive index of air as 1) and n2 = 1.66 (refractive index of the plastic light pipe), we can rearrange the equation to solve for sin(theta1):
sin(theta1) = n2 / n1
sin(theta1) = 1.66 / 1
sin(theta1) = 1.66
However, the sine of an angle cannot be greater than 1. Therefore, there is no critical angle for total internal reflection when light travels from air to the plastic light pipe. Total internal reflection does not occur in this case.
(b) Water as the initial medium:
Given n1 = 1.33 (refractive index of water) and n2 = 1.66 (refractive index of the plastic light pipe), we can use the same equation to find sin(theta1):
sin(theta1) = n2 / n1
sin(theta1) = 1.66 / 1.33
sin(theta1) ≈ 1.248
To find the angle theta1, we can take the inverse sine of sin(theta1):
theta1 = arcsin(sin(theta1))
theta1 ≈ arcsin(1.248)
However, since the sine of an angle cannot exceed 1, there is no real solution for theta1 in this case. Total internal reflection does not occur when light travels from water to the plastic light pipe.
Therefore, for both (a) air and (b) water as the initial medium, total internal reflection does not occur when light enters the plastic light pipe with a refractive index of 1.66.
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Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (n core
=1.519) and the cladding (n cladding
=1.429). A 50% Part (a) Numerically, what is the largest angle (in degrees) a ray will make with respect to the interface internal reflection? θ max
= Hints: deduction per hint. Hints remaining: 2
deduction per feedback. (4 50% Part (b) Suppose you wanted the largest angle at which total internal reflection occurred to be θ max
=5 (6\%) Problem 6: Suppose a 200-mm focal length telephoto lens is being used to photograph mountains 9.5 km away. ( 50% Part (a) What is image distance, in meters, for this lens? d i
= \begin{tabular}{llll} \hline Hints: deduction per hint. Hints remaining: 1 & Feedback: \end{tabular}
This makes the critical angle 5 degrees. To prove this, we use the same formula:sinθc = n2/n1sin(5) = 1.054/1.519θc = 5 degrees
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore=1.519) and the cladding (ncladding=1.429).A 50%Part
(a) Numerically, what is the largest angle (in degrees) a ray will make with respect to the interface internal reflection? θmax=In order to determine the angle that a ray will make with respect to the interface internal reflection, we use Snell's Law: n1sinθ1 = n2sinθ2
where:n1 is the refractive index of the medium the ray is coming fromθ1 is the angle of incidence measured from the normaln2 is the refractive index of the medium the ray is enteringθ2 is the angle of refraction measured from the normalWhen light travels from a medium of a higher refractive index to one of a lower refractive index (i.e. from the core to the cladding),
the angle of refraction is larger than the angle of incidence; that is, the ray is refracted away from the normal. At the critical angle, however, the angle of refraction is 90 degrees. Thus, sinθ2 = 1. Setting sinθ1 = n2/n1, we get the critical angle formula:sinθc = n2/n1θc = sin^(-1)(n2/n1)
The maximum angle a ray will make with respect to the interface internal reflection will be the complement of the critical angle:θmax = 90 - θc = 90 - sin^(-1)(n2/n1) = 90 - sin^(-1)(1.429/1.519) = 42.45 degrees50%Part (b) Suppose you wanted the largest angle at which total internal reflection occurred to be θmax=5°. You could achieve this by decreasing the refractive index of the cladding to ncladding = 1.054.
This makes the critical angle 5 degrees. To prove this, we use the same formula:sinθc = n2/n1sin(5) = 1.054/1.519θc = 5 degrees
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A closed and elevated vertical cylindrical tank with diameter 2.20 m contains water to a depth of 0.900 m . A worker accidently pokes a circular hole with diameter 0.0190 m in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of 5.00×103Pa at the surface of the water. Ignore any effects of viscosity.
The rate at which water flows out of the hole in the tank is approximately 1.51×[tex]10^{-3}[/tex] cubic meters per second.
To determine the rate at which water flows out of the hole in the tank, we can apply Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a flowing system.
First, let's find the velocity of the water flowing out of the hole.
The gauge pressure at the surface of the water is given as 5.00×10^3 Pa.
We can assume atmospheric pressure at the hole, so the total pressure at the hole is the sum of the gauge pressure and atmospheric pressure, which is 5.00×[tex]10^3[/tex] Pa + 1.01×[tex]10^5[/tex] Pa = 1.06×[tex]10^5[/tex] Pa.
According to Bernoulli's equation, the total pressure at the hole is equal to the pressure due to the water column plus the dynamic pressure of the flowing water:
P_total = P_water + (1/2)ρ[tex]v^2[/tex] + P_atm,
where P_total is the total pressure, P_water is the pressure due to the water column, ρ is the density of water, v is the velocity of the water flowing out of the hole, and P_atm is atmospheric pressure.
Since the tank is vertically oriented and the hole is at the bottom, the pressure due to the water column is ρgh, where h is the height of the water column above the hole. In this case, h = 0.900 m.
We can rewrite Bernoulli's equation as:
P_total = ρgh + (1/2)ρ[tex]v^2[/tex] + P_atm.
Now we can solve for v. Rearranging the equation, we get:
(1/2)ρ[tex]v^2[/tex] = P_total - ρgh - P_atm,
[tex]v^2[/tex] = 2(P_total - ρgh - P_atm)/ρ,
v = [tex]\sqrt[/tex](2(P_total - ρgh - P_atm)/ρ).
Now we can plug in the known values:
P_total = 1.06×[tex]10^5[/tex] Pa,
ρ = 1000 kg/[tex]m^3[/tex] (density of water),
g = 9.81 m/[tex]s^2[/tex] (acceleration due to gravity),
h = 0.900 m,
P_atm = 1.01×[tex]10^5[/tex] Pa (atmospheric pressure).
Substituting these values into the equation, we can calculate the velocity v of the water flowing out of the hole.
After finding the velocity, we can then calculate the rate at which water flows out of the hole using the equation for the volume flow rate:
Q = Av,
where Q is the volume flow rate, A is the cross-sectional area of the hole (π[tex]r^2[/tex], where r is the radius of the hole), and v is the velocity of the water.
Let's substitute the known values into the equations to calculate the velocity and volume flow rate.
First, let's calculate the velocity:
v =[tex]\sqrt[/tex](2(P_total - ρgh - P_atm)/ρ)
= [tex]\sqrt[/tex](2((1.06×10^5 Pa) - (1000 kg/m^3)(9.81 m/s^2)(0.900 m) - (1.01×10^5 Pa))/(1000 kg/m^3))
Simplifying the equation:
v ≈ 5.32 m/s
Next, let's calculate the cross-sectional area of the hole:
A = πr^2
= π(0.0190 m/2)^2
Simplifying the equation:
A ≈ 2.84×[tex]10^{-4}[/tex] [tex]m^2[/tex]
Finally, let's calculate the volume flow rate:
Q = Av
= (2.84×[tex]10^{-4}[/tex] [tex]m^2[/tex])(5.32 m/s)
Simplifying the equation:
Q ≈ 1.51×[tex]10^{-3}[/tex] [tex]m^3[/tex]/s
Therefore, the rate at which water flows out of the hole in the tank is approximately 1.51×[tex]10^{-3}[/tex] cubic meters per second.
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Two stationary point charges experience a mutual electric force of magnitude 108 N. Subsequently, if the distance between the two point charges is tripled while the magnitude of both charges is cut in half.
What is the magnitude of the resultant electric force on either charge?
a. 6.0 N
b. 3.0 N
c. 12 N
d. 9.0 N
e. 27 N
The correct answer the magnitude of the resultant electric force on either charge is Option d.9.0 N
Let the original magnitude of one charge be q1 and the original magnitude of the other charge be q2. The original distance between the two charges is r.
The magnitude of the force between two point charges q1 and q2 is given by Coulomb's law as:F=kq1q2/r²Where k is Coulomb's constant which is 9 × 10^9 Nm²/C².Subsequently, if the distance between the two point charges is tripled while the magnitude of both charges is cut in half, the new distance between the two charges is 3r and the new magnitude of both charges is (1/2)q.
The force between the two charges with the new conditions is given by:F'=k((1/2)q)(1/2)q/(3r)²F'=kq²/27r²Since the magnitude of the force between two stationary point charges is the same for each charge, the magnitude of the resultant electric force on either charge is given by:F''=F'/2F''=kq²/54r²The ratio of the new force to the old force is:F''/F=kq/108r².
The magnitude of the force on each charge is:F1=F2=F''/2F1=F2=kq²/108r²The magnitude of the force on each charge is kq²/108r². Answer: d. 9.0 N.
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Calculate the maximum kinetic energy of a beta particle when 19K decays via 3.
The Q-value of the decay is 21.46 MeV.The electron binding energy of 19Ca is 3.210 MeV. Therefore, the maximum kinetic energy of the beta particle is:Kmax = Q – EbKmax = 21.46 MeV – 3.210 MeVKmax = 18.25 MeV
When 19K decays to 19Ca via β− decay, the maximum kinetic energy of the beta particle can be calculated by using the following formula: Kmax = Q – Eb Here, Kmax is the maximum kinetic energy of the beta particle, Q is the Q-value of the decay, and Eb is the electron binding energy of the 19Ca atom.
The Q-value of the decay can be calculated using the mass-energy balance equation.
This equation is given by:m(19K)c² = m(19Ca)c² + melectronc² + QHere, melectronc² is the rest mass energy of the electron, which is equal to 0.511 MeV/c².
Substituting the atomic masses from the periodic table, we get:m(19K) = 18.998 403 163 u, m(19Ca) = 18.973 847 u.
Substituting these values into the equation and simplifying, we get:Q = [m(19K) – m(19Ca) – melectron]c²Q = [18.998 403 163 u – 18.973 847 u – 0.000 548 579 u] × (931.5 MeV/u)Q = 0.023 007 u × (931.5 MeV/u)Q = 21.46 MeV
Therefore, the Q-value of the decay is 21.46 MeV. The electron binding energy of 19Ca is 3.210 MeV. Therefore, the maximum kinetic energy of the beta particle is: Kmax = Q – EbKmax = 21.46 MeV – 3.210 MeVKmax = 18.25 MeV
Therefore, the maximum kinetic energy of the beta particle is 18.25 MeV.
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The three lines on the distance-time graph in Figure represent the motion of three objects: (a) Which object has travelled farthest at time t=5 s ? (b) How far has each object travelled at time t=3 s? (c) What is the slope of each line?
(a) To determine which object has traveled farthest at time t = 5 s. (b) To find the distance traveled by each object at time t = 3 s. (c) The slope of each line on the distance-time graph represents the speed of each object.
(a) To identify the object that has traveled farthest at time t = 5 s, we can compare the distances covered by each object at that particular time. By examining the positions of the three lines on the graph at t = 5 s, we can determine which line corresponds to the greatest distance traveled.
(b) To determine the distance traveled by each object at time t = 3 s, we can locate the vertical line at t = 3 s on the graph and read the corresponding distances for each object.
(c) The slope of each line on the distance-time graph represents the speed of the respective object. The steeper the slope, the greater the speed.
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An electron (mass 9 x 10⁻³¹ kg) is traveling at a speed of 0.91c in an electron accelerator. An electric force of 1.6 x 10 N is applied in the direction of motion while the electron travels a distance of 2 m. You need to find the new speed of the electron. Which of the following steps must be included in your solution to this problem? (a) Calculate the initial particle energy Yimc of the electron. (b) Calculate the final particle energy y&mc? of the electron. (c) Determine how much time it takes to move this distance. (d) Use the expression m[512 to find the kinetic energy of the electron. (e) Calculate the net work done on the electron. (f) Use the final energy of the electron to find its final speed. What is the new speed of the electron as a fraction of c?
The new speed of the electron as a fraction of c is 0.9655.
Mass of electron = m = 9 x 10⁻³¹ kg
Speed of electron = u = 0.91c
Electric force = F = 1.6 x 10 N
Crossing distance = s = 2 m
Electric force = F = ma
where, F = Electric force, m = Mass of the electron, a = Acceleration of the electron.
Using above equation, we get, a = F/ma = F/m = 1.6 x 10 / 9 x 10⁻³¹ a = 1.78 x 10⁴ m/s²
Now, we can calculate the time taken by electron to travel a distance of 2m using s = ut + ½ at²
where, u = Initial speed of electron, t = Time taken by electron to travel distance s, a = Acceleration of electron, s = Distance travelled by electron.
So, t = s / (u/2 + ½ a)
We get, t = 2 / [0.91c/2 + 1/2 * 1.78 x 10⁴]
= 5.71 x 10⁻¹⁰ s
Kinetic energy = [m / √(1- (v/c)²)] c² - mc²
where, Kinetic energy = Final kinetic energy of electron, m = Mass of the electron, v = Final speed of the electron.
So, K.E = [9 x 10⁻³¹ / √(1-(v/c)²)] c² - (9 x 10⁻³¹) c²
Now, calculate the net work done on the electron. Wnet = K.E - K.Eo
where, Wnet = Net work done on electron, K.E = Final kinetic energy of electron, K.Eo = Initial kinetic energy of electron.
K.Eo = [9 x 10⁻³¹ / √(1-(u/c)²)] c² - (9 x 10⁻³¹) c²
we get, Wnet = [9 x 10⁻³¹ / √(1-(v/c)²)] c² - [9 x 10⁻³¹ / √(1-(u/c)²)] c²
Simplify this expression, Wnet = 0.5 x 9 x 10⁻³¹ [(1/√(1-(v/c)²)] c² - [(1/√(1-(u/c)²)] c²
= 0.5 x m [(1/√(1-(v/c)²)] c² - [(1/√(1-(u/c)²)] c²
Finally, use the work-energy principle. We know that, Wnet = ΔK.E
Wnet = Work done on the particle, ΔK.E = Change in kinetic energy of the particle.
Since the electron is being accelerated, the force acting on it is in the same direction as the direction of motion and hence, the work done is positive. So, we can write Wnet = K.E - K.Eo.
Now, put the values of Wnet, ΔK.E, K.E and K.Eo, we get,0.5 x m [(1/√(1-(v/c)²)] c² - [(1/√(1-(u/c)²)] c²
= [(9 x 10⁻³¹ / √(1-(v/c)²)] c² - [(9 x 10⁻³¹ / √(1-(u/c)²)] c² - [(9 x 10⁻³¹ / √(1-(u/c)²)] c²
Now, we can calculate the final kinetic energy of the electron, Kinetic energy = (Wnet + K.Eo)K.E = 0.5 x m [(1/√(1-(v/c)²)] c² + [(9 x 10⁻³¹ / √(1-(u/c)²)] c²K.E
= [9 x 10⁻³¹ / √(1-(v/c)²)] c²v/c
= √[1 - ((m/m+1)(c/u²t²))]v/c
= √[1 - ((9 x 10⁻³¹/10⁻³¹ + 1)(3 x 10⁸/(0.91 x 3 x 10⁸)² x (5.71 x 10⁻¹⁰)²))]v/c = 0.9655
Therefore, the new speed of the electron as a fraction of c is 0.9655.
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Impulse has the same SI units as work linear momentum kinetic energy all of the above Question 3 (1 point) ✓ Saved Momentum is conserved when An insect collides with the windshield of a moving car. An electron splits an atom into many subatomic particles. A rifle fires a bullet and the gun recoils. all of the above Choose the correct statement. Work is a vector quantity. Work is not a scalar quantity. W=FΔdcosθ
W=Fp
Choice D, all of the above, is the correct answer. For the third question, the correct statement is: W = FΔd cosθ.Work is a scalar quantity that represents the transfer of energy that occurs when a force is applied to an object and it moves through a distance.
Impulse has the same SI units as momentum. Impulse and momentum share the same SI units, which are kg m/s. Impulse and momentum are also related to each other. Impulse is defined as the change in momentum of an object. Impulse = Δp = mΔvMomentum = p = mvwhere m is the mass of the object and v is its velocity.Work, linear momentum, and kinetic energy are not equivalent to impulse. They have different SI units and meanings.Work is the transfer of energy that occurs when a force is applied to an object and it moves through a distance. Its SI units are joules (J).Linear momentum is the product of an object's mass and velocity. Its SI units are kg m/s.Kinetic energy is the energy an object has due to its motion. Its SI units are also joules (J).For the second question, momentum is conserved when an insect collides with the windshield of a moving car, an electron splits an atom into many subatomic particles, a rifle fires a bullet and the gun recoils. Choice D, all of the above, is the correct answer. For the third question, the correct statement is: W = FΔd cosθ.Work is a scalar quantity that represents the transfer of energy that occurs when a force is applied to an object and it moves through a distance. It is calculated using the formula W = FΔd cosθ, where F is the force applied, Δd is the displacement of the object, and θ is the angle between the force and the displacement.
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An ocean-going research submarine has a 20-cm-diameter window 8.0 cm thick. The manufacturer says the window can withstand forces up to 1.0 X 100 N. What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1.0 atm.
The maximum safe depth of the submarine is approximately 10,317 meters can be determined by calculating the pressure exerted on the window and comparing it to the manufacturer's stated limit.
To calculate the maximum safe depth of the submarine, we need to consider the pressure exerted on the window. The pressure exerted by a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the fluid is seawater.
First, we need to determine the pressure exerted on the window at the maximum safe depth. The pressure inside the submarine is maintained at 1.0 atm, which is equivalent to 101,325 Pa. We can assume that the density of seawater is approximately [tex]1,030 kg/m^3[/tex] and the acceleration due to gravity is [tex]9.8 m/s^2[/tex].
Using the equation P = ρgh, we can rearrange it to solve for h: h = P / (ρg). Plugging in the values, we have h = [tex]101,325 Pa / (1,030 kg/m^3 * 9.8 m/s^2)[/tex], which gives us the maximum safe depth of the submarine.
To find out the numerical value, we need to evaluate the expression. The maximum safe depth of the submarine is approximately 10,317 meters.
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You are spending the summer as an assistant learning how to navigate on a large ship carrying freight across Lake Erie. One day, you and your ship are to travel across the lake a distance of 200 km traveling due north from your origin port to your destination port. Just as you leave your origin port, the navigation electronics go down. The cap- tain continues sailing, claiming he can depend on his years of experience on the water as a guide. The engineers work on the navigation system while the ship continues to sail, and winds and waves push it off course. Eventually, enough of the navigation system comes back up to tell you your location. The system tells you that your current position is 50.0 km north of the origin port and 25.0 km east of the port. The captain is a little embarrassed that his ship is so far off course and barks an order to you to tell him immedi- ately what heading he should set from your current position to the destination port. Give him an appropriate heading angle.
You should advise the captain to set a heading angle of approximately 63.43 degrees from your current position towards the destination port.
To determine the heading angle from your current position to the destination port, you can use trigonometry. Given that your current position is 50.0 km north and 25.0 km east of the origin port, you can consider these values as the lengths of the legs of a right triangle.
The desired heading angle can be found using the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right triangle. In this case, the opposite side is the northward distance (50.0 km) and the adjacent side is the eastward distance (25.0 km).
The heading angle (θ) can be calculated as:
θ = tan^(-1)(opposite/adjacent)
θ = tan^(-1)(50.0 km/25.0 km)
Using a calculator, the approximate value of the heading angle is:
θ ≈ 63.43 degrees
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a) Calculate the wavelength of light emitted by a Hydrogen atom when its electron decays from the n=3 to the n=1 state energy level. b) With respect to the photoelectric effect, the work function of Lead ( Pb) is 4.25eV. What is the cut-off wavelength of Pb ? c) A sample of Pb is illuminated with light having the wavelength calculated in part a). Calculate the velocity of the emitted electrons.
a) When an electron in a hydrogen atom transitions from the n=3 to the n=1 energy level, the wavelength of light emitted can be calculated using the Rydberg formula: 1/λ = R_H * (1/n_1^2 - 1/n_2^2), where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m^-1), n_1 is the initial energy level (n=3), and n_2 is the final energy level (n=1).
b) The cut-off wavelength of lead (Pb) can be determined based on the work function, which is the minimum energy required to remove an electron from the metal surface. The relationship between the cut-off wavelength (λ_cutoff) and the work function (Φ) is given by λ_cutoff = hc / Φ, where h is Planck's constant (approximately 6.626 × 10^-34 J·s) and c is the speed of light (approximately 3.00 × 10^8 m/s). By substituting the value of the work function (4.25 eV) into the equation, we can calculate the cut-off wavelength of lead.
c) Once the wavelength of the emitted light from part a) is known, the velocity of the emitted electrons can be determined using the de Broglie wavelength equation: λ = h / mv, where m is the mass of the electron and v is its velocity. By rearranging the equation, we can solve for the velocity: v = h / (mλ). By substituting the mass of an electron and the calculated wavelength into the equation, we can find the velocity of the emitted electrons.
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A deuteron, consisting of a proton and neutron and having mass 3.34 x 10⁻²⁷ kg, is traveling at 0.942c relative the Earth in a linear accelerator. Calculate the deuteron's rest energy, v-factor, total energy, and kinetic energy. (a) rest energy (Give your answer to at least three significant figures.) _______________ J
(b) y-factor ___________
(c) total energy
_______________ J
(d) kinetic energy
_______________ J
A deuteron, with proton and neutron having mass 3.34 x 10⁻²⁷ kg, is traveling at 0.942c relative the Earth in a linear accelerator, then it's Rest energy = 3.009 x 10⁻¹⁰ J, v-factor = 0.942, Total energy = 2.643 x 10⁻¹⁰ J, Kinetic energy = -3.66 x 10⁻¹¹ J.
It is given that, Mass of the deuteron (m) = 3.34 x 10⁻²⁷ kg, Speed of light (c) = 3 x 10^8 m/s, Speed of the deuteron (v) = 0.942c.
(a) Rest Energy:
E_rest = m * c²
E_rest = (3.34 x 10⁻²⁷ kg) * (3 x 10⁸ m/s)²
E_rest = 3.009 x 10⁻¹⁰ J
(b) v-factor:
β = v / c
β = (0.942c) / (3 x 10⁸ m/s)
β = 0.942
(c) Total Energy:
To find the total energy, we need to calculate the γ factor (gamma) using the v-factor (β):
γ = 1 / sqrt(1 - β²)
γ = 1 / sqrt(1 - (0.942)²)
γ = 2.943
Now we can calculate the total energy:
E_total = γ * m * c²
E_total = (2.943) * (3.34 x 10⁻²⁷ kg) * (3 x 10⁸ m/s)²
E_total = 2.643 x 10⁻¹⁰ J
(d) Kinetic Energy:
To calculate the kinetic energy, we subtract the rest energy from the total energy:
E_kinetic = E_total - E_rest
E_kinetic = (2.643 x 10⁻¹⁰ J) - (3.009 x 10⁻¹⁰ J)
E_kinetic = -3.66 x 10⁻¹¹ J
The negative sign indicates that the kinetic energy is negative, which means the deuteron is moving at a speed below its rest frame.
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Mod1HW
Problem 20: A student begins at rest and then walks north at a speed of v1 = 0.75 m/s. The student then turns south and walks at a speed of v2 = 0.76 m/s. Take north to be the positive direction. Refer to the figure.
Part (a) What is the student's overall average velocity vavg, in meters per second, for the trip assuming the student spent equal times at speeds v1 and v2?
Part (b) If the student travels in the stated directions for 30.0 seconds at speed v1 and for 20.0 seconds at speed v2, what is the net displacement, in meters, during the trip?
Part (c) If it takes the student 5.0 s to reach the speed v1 from rest, what is the magnitude of the student’s average acceleration, in meters per second squared, during that time?
Part a)The total distance covered is, D = 2dThe average velocity is given byvavg = D / ttotal= 2d / (2d / v1 + d / v2)= (2v1v2) / (v1 + v2)= (2 × 0.75 × 0.76) / (0.75 + 0.76)≈ 0.757 m/s.Part b)The net displacement is given byx = d1 - d2= 22.5 - 15.2= 7.3 m. Partc).The magnitude of the student's average acceleration during that time is 0.15 m/s².
Part a) Let the distance traveled in each direction be d.The time taken to travel in each direction is given by:t = d / v1 for the northward directiont = d / v2 for the southward direction.The total time taken is, ttotal = 2t = 2d / v1 + v2The total distance covered is, D = 2dThe average velocity is given byvavg = D / ttotal= 2d / (2d / v1 + d / v2)= (2v1v2) / (v1 + v2)= (2 × 0.75 × 0.76) / (0.75 + 0.76)≈ 0.757 m/s.
Part b)The distance covered in each direction is given byd1 = v1t1 = 0.75 × 30 = 22.5 md2 = v2t2 = 0.76 × 20 = 15.2 mThe net displacement is given byx = d1 - d2= 22.5 - 15.2= 7.3 m.
Part c)Initial velocity, u = 0; Final velocity, v = v1 = 0.75 m/sThe time taken to reach the final velocity is, t = 5 s Average acceleration is given byaavg = (v - u) / t= 0.75 / 5= 0.15 m/s²Therefore, the magnitude of the student's average acceleration during that time is 0.15 m/s². The answer is 0.15 m/s².
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The student's overall average velocity for the trip is zero. The net displacement during the trip is 7.3 meters. The magnitude of the student's average acceleration during the time it took to reach speed v1 from rest is 0.15 m/s².
Explanation:Part (a): To find the student's overall average velocity, we can use the formula average velocity = total displacement / total time. Since the student spent equal times at speeds v1 and v2, the total displacement is zero. Therefore, the overall average velocity is also zero.
Part (b): To find the net displacement, we need to calculate the distance traveled at each speed and the direction. In the north direction, the student travels for 30.0 seconds at a speed of v1 = 0.75 m/s, so the northward displacement is 30.0 s × 0.75 m/s = 22.5 m. In the south direction, the student travels for 20.0 seconds at a speed of v2 = 0.76 m/s, so the southward displacement is 20.0 s × (-0.76 m/s) = -15.2 m. The net displacement is the sum of the displacements, which is 22.5 m - 15.2 m = 7.3 m.
Part (c): Average acceleration is given by the formula average acceleration = final velocity - initial velocity / time taken. The initial velocity is 0 m/s, the final velocity is 0.75 m/s, and the time taken is 5.0 seconds. Plugging in these values, we get average acceleration = (0.75 m/s - 0 m/s) / 5.0 s = 0.15 m/s².
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A circuit connects battery to three light bulbs in parallel. In other words, all the light bulbs are in parallel with one another, and in parallel with the battery. What happens to the circuit if one of the light bulb burns out? Why? A. Total resistance increases, other bulbs get brighter B. Total resistance increases, other bulbs get dimmer C. Total resistance increases, brightness of other bulbs does not change D. All the bulbs go out E. Total resistance decreases, other bulbs get brighter F. Total resistance decreases, other bulbs get dimmer G. Total resistance decreases, brightness of other bulbs does not change
If one of the light bulb burns out, Total resistance increases, other bulbs get dimmer. The circuit would not be broken if one of the bulbs burns out. This is the effect of a parallel circuit when one component fails. Therefore. the correct answer is option B.
In a parallel circuit, each device operates independently. As a result, if one component fails, it does not cause the others to stop working. However, since the resistance of each bulb is fixed, the total resistance of the circuit decreases as bulbs are added.
When a bulb burns out, the resistance of the circuit rises, making the other bulbs dimmer. Because the current in a parallel circuit is divided among the components, the current flowing through each remaining bulb would decrease if one bulb burns out.
So, if one bulb fails, the voltage across it would drop, and it would get dimmer. That's why in parallel circuit the bulbs are installed in parallel to ensure that they function independently of each other. So, option B is the correct answer.
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1.50 moles of a monatomic ideal gas goes isothermally from state 1 to state 2. P1 = 3.6x10⁵ Pa, V1 = 60 m³, and P2 = 5.8 x 10⁵ Pa. What is the volume in state 2, in m³? Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement.
The volume in state 2 of an isothermal process, with initial pressure of 3.6 x 10⁵ Pa and volume of 60 m³, is 216 m³. The answer is rounded to 2 significant figures.
To find the volume in state 2, we can use the ideal gas law equation:
P₁V₁ = P₂V₂,
where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume, respectively.
Given:
P₁ = 3.6 x 10⁵ Pa,
V₁ = 60 m³,
P₂ = 5.8 x 10⁵ Pa.
Rearranging the equation and solving for V₂:
V₂ = (P₁ * V₁) / P₂.
Substituting the values:
V₂ = (3.6 x 10⁵ Pa * 60 m³) / (5.8 x 10⁵ Pa).
Calculating V₂:
V₂ = 216 m³.
Therefore, the volume in state 2 is 216 m³ (rounded to 2 significant figures).
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The correct answer is: A,Aω,Aω2 The position of an object moving in simple harmonic motion is given by the equation x(t)=Asin(ωt+θ), where A=−3.7 m, at=2.0rad/s and θ=0.20rad. What is the speed of the object when it is at x=−1.5 m ? Select one: a. 7.0 m/s b. 6.8 m/s c. 3.8 m/s d. 3.4 m/s Take the denvative of x(t) to find the velocity as a function of tate: x(t)=Asin(ωt+θ)v(t)=dtdx
The speed of the object when it is at x = -1.5 m is 7.0 m/s. Answer: a. 7.0 m/s.
Given data,A = -3.7 mω = 2.0 rad/st = ?θ = 0.20 radWe know that velocity as a function of time is given by the derivative of position as a function of time, that is,v(t) = d/dt [x(t)]v(t) = d/dt [Asin(ωt + θ)]v(t) = Aω cos(ωt + θ)Now, the position of the object is given byx(t) = Asin(ωt + θ)Now, substituting the given values, we getx(t) = -3.7 sin(2t + 0.20) mNow, the object is at x = -1.5 mHence, -1.5 = -3.7 sin(2t + 0.20)Solving for t, we gett = 0.835 sNow, substituting t = 0.835 s in the equation of velocity as a function of time, we getv(t) = Aω cos(ωt + θ)v(t) = -3.7 × 2.0 cos(2(0.835) + 0.20) m/sv(t) = -7.0 m/sTherefore, the speed of the object when it is at x = -1.5 m is 7.0 m/s. Answer: a. 7.0 m/s.
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Suppose you try to cool the kitchen of your house by leaving the refrigerator door open. What happens? Why? Would the result be the same if you left open a picnic cooler full of ice? Explain the reason for any differences.
Is it a violation of the second law of thermodynamics to convert mechanical energy completely into heat? To convert heat completely into work? Explain your answers.
Real heat engines, like the gasoline engine in a car, always have some friction between their moving parts, although lubricants keep the friction to a minimum. Would a heat engine with completely frictionless parts be 100% efficient? Why or why not? Does the answer depend on whether or not the engine runs on the Carnot cycle? Again, why or why not?
A heat engine with completely frictionless parts would still not be 100% efficient even if it ran on the Carnot cycle.
Suppose you try to cool the kitchen of your house by leaving the refrigerator door open. What happens? Why?Would the result be the same if you left open a picnic cooler full of ice? Explain the reason for any differences.If you leave the refrigerator door open, the room may become slightly colder initially, but the overall effect will be to warm up the room. This is because the refrigerator will work to cool down the air inside it but at the same time will pump the heat out into the room. As a result, the room’s temperature will rise. If you left a picnic cooler full of ice open in the room, the ice would eventually melt and the water would eventually warm up to room temperature, raising the temperature of the room.
However, the cooling effect of the ice will be greater than the heating effect of the air that escapes. Therefore, it will be more efficient in cooling the room for a shorter time.Is it a violation of the second law of thermodynamics to convert mechanical energy completely into heat? To convert heat completely into work? Explain your answers.No, it is not a violation of the second law of thermodynamics to convert mechanical energy completely into heat because heat is a form of energy, and the second law of thermodynamics states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another.
However, it is impossible to convert heat completely into work because some heat energy will always be lost to the environment, and the second law of thermodynamics prohibits the conversion of heat energy completely into work.Real heat engines, like the gasoline engine in a car, always have some friction between their moving parts, although lubricants keep the friction to a minimum. Would a heat engine with completely frictionless parts be 100% efficient? Why or why not? Does the answer depend on whether or not the engine runs on the Carnot cycle?
Again, why or why not?A heat engine with completely frictionless parts would not be 100% efficient because some energy would still be lost as heat due to the second law of thermodynamics. The answer does not depend on whether or not the engine runs on the Carnot cycle because the Carnot cycle assumes an ideal engine with no friction, which is not possible in the real world. Therefore, a heat engine with completely frictionless parts would still not be 100% efficient even if it ran on the Carnot cycle.
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a) At what frequency would a 6.0 mH inductor and a 10 nF capacitor have the same reactance? (b) What would the reactance be? (©) Show that this frequency would be the nat- ural frequency of an oscillating circuit with the same L and C.
Answer:
The frequency at which the 6.0 mH inductor and 10 nF capacitor have the same reactance is approximately 20,462 Hz.
Reactance of an inductor (XL) is given by:
XL = 2πfL
Reactance of a capacitor (XC) is given by:
XC = 1 / (2πfC)
Where f is the frequency, L is the inductance, and C is the capacitance.
Setting XL equal to XC:
2πfL = 1 / (2πfC)
Simplifying the equation:
f = 1 / (2π√(LC))
L = 6.0 mH
= 6.0 x 10^(-3) H
C = 10 nF
= 10 x 10^(-9) F
Substituting the given values into the equation:
f = 1 / (2π√(6.0 x 10^(-3) H * 10 x 10^(-9) F))
Simplifying the expression:
f = 1 / (2π√(60 x 10^(-12) H·F))
f = 1 / (2π√(60 x 10^(-12) s^2 / C^2))
f = 1 / (2π x 7.75 x 10^(-6) s)
f ≈ 20,462 Hz
Therefore, the frequency at which the 6.0 mH inductor and 10 nF capacitor have the same reactance is approximately 20,462 Hz.
To show that this frequency is the natural frequency of an oscillating circuit with the same L and C, we can use the formula for the natural frequency of an LC circuit:
fn = 1 / (2π√(LC))
Substituting the given values into the formula:
fn = 1 / (2π√(6.0 x 10^(-3) H * 10 x 10^(-9) F))
fn = 1 / (2π√(60 x 10^(-12) H·F))
fn = 1 / (2π√(60 x 10^(-12) s^2 / C^2))
fn = 1 / (2π x 7.75 x 10^(-6) s)
fn ≈ 20,462 Hz
We can see that this frequency matches the frequency obtained earlier, confirming that it is the natural frequency of an oscillating circuit with the same L and C.
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What do you need to find the intensity of an electromagnetic wave?
Both the electric and magnetic field strengths.
Either the electric or magnetic field strength.
Only the electric field strength.
Only the magnetic field strength.
To find the intensity of an electromagnetic wave, we need to know the electric and magnetic field strengths as they are interdependent. The correct option is 1) Both the electric and magnetic field strengths.
The intensity of an electromagnetic wave is given by the energy transferred per unit area per unit time and is proportional to the square of the electric and magnetic field strengths. Therefore, if either the electric or magnetic field strength is missing, it will be impossible to determine the intensity accurately. The electric and magnetic fields oscillate perpendicular to each other and the direction of propagation of the wave. They have the same amplitude, frequency, and wavelength, but they differ in phase.
The intensity of an electromagnetic wave can also be determined by measuring the average power per unit area over a period. In summary, both electric and magnetic field strengths are required to calculate the intensity of an electromagnetic wave accurately. It is important to note that these fields are interdependent on each other, and a change in one can affect the other. Therefore, accurate measurements are crucial in the determination of the intensity of electromagnetic waves.
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Two trains are traveling toward each other at 30.9 m/s relative to the ground. One train is blowing a whistle at 510 Hz. (Give your answers to at least three significant figures.) (a) What frequency will be heard on the other train in still air? Hz (b) What frequency will be heard on the other train if the wind is blowing at 30.9 m/s toward the whistle and away from the listener? Hz (c) What frequency will be heard if the wind direction is reversed? Hz
(a) The frequency heard on the other train in still air will be 510 Hz.
(b) The frequency heard on the other train, with the wind blowing toward the whistle and away from the listener, will be higher than 510 Hz.
(c) The frequency heard on the other train, with the wind direction reversed, will be lower than 510 Hz.
(a) When two trains approach each other, the frequency heard on the other train in still air is the same as the emitted frequency, which is 510 Hz in this case. This is because the speed of sound is the same in both directions relative to the ground.
(b) When the wind is blowing at 30.9 m/s toward the whistle and away from the listener, the effective speed of sound is increased. This is due to the additive effect of the wind speed to the speed of sound. As a result, the frequency heard on the other train will be higher than the emitted frequency of 510 Hz.
(c) Conversely, when the wind direction is reversed, the effective speed of sound is reduced. The wind speed is subtracted from the speed of sound, leading to a lower effective speed of sound. Therefore, the frequency heard on the other train will be lower than 510 Hz.
These changes in frequency, known as the Doppler effect, occur due to the relative motion between the source (train) and the observer (other train) as well as the medium through which the sound waves travel (air).
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A string in a guitar (string instrument) is 2.4m long, and the speed of sound along this string is 450m/s. Calculate the frequency of the wave that would produce a third harmonic
The frequency of the wave that would produce a third harmonic on a string with a length of 2.4 m and a speed of sound of 450 m/s is approximately 281.25 Hz.
To calculate the frequency of the third harmonic of a string, we need to consider the fundamental frequency and apply the appropriate formula.
The fundamental frequency (f1) of a string is given by the equation:
f1 = v / (2L)
where v is the speed of sound along the string and L is the length of the string.
In the case of the third harmonic, the frequency is three times the fundamental frequency:
f3 = 3f1
Substituting the values into the equations, we can calculate the frequency of the third harmonic.
f1 = 450 m/s / (2 * 2.4 m)
f1 ≈ 93.75 Hz
f3 = 3 * 93.75 Hz
f3 ≈ 281.25 Hz
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Electric force \& electric potentials For ench electrostatic figure circle A or B. Charges are explicit in Q17, 21 \& mplicit in Q18-20 If you choose B then you MUSI explain why the lines shown ate not electric field lines. 17. Simple ForcePotential Question A. This could be an Electric Field. B. This is NOT an Electrie Field because: 18. Simple Force Potential Question A. This coud be an Electric Field. B. This is NOT an Electric Field becmase: 19. Simple Force.Porential Question A. This could be an Electnc Field. B. This in NOT an Electric Field because: 20. Simple Force Potential Question A. This could be an Electne Freld. B. This is NOT an Electric Field becatise: 21. Simple ForcePotential Question A. This could be an Electric Field. B. This is NOI an Electric Field because:
This could be an Electric Field. B. This is NOT an Electric Field because: There is no charge represented in the figure; hence, it does not represent an electric field.
The electric force, as well as electric potentials, is given by Coulomb's law. Coulomb's law states that electric force between two charges, Q1 and Q2 is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The charges in this question are explicit in Q17, 21 & implicit in Q18-20. Let's discuss the circles. Circles A and B are simple force-potential figures. Circle A is a graphical representation of electric field lines. This is because the arrows show the direction of force that would be exerted on a unit charge at every point, and the density of lines indicates the strength of the electric field.
On the other hand, circle B shows equipotential lines. This is because the lines are parallel to each other and the potential difference between them is constant. If circle B showed electric field lines, the arrows would be perpendicular to the equipotential lines, whereas in this figure, the lines are not perpendicular. Hence, the lines in circle B are not electric field lines.
It is essential to understand that equipotential lines always cross at right angles. Circle A: 17. Simple Force Potential Question A. This could be an Electric Field. B. This is an Electrie Field because: It is a typical electric field with its field lines emerging from the positive charges and terminating at the negative charges. Circle B: 18. Simple Force Potential Question A.
This could be an Electric Field. B. This is NOT an Electric Field because: The parallel lines in the graph indicate equipotential lines and not electric field lines. Circle A: 19. Simple Force Potential Question A. This could be an Electnc Field. B. This is NOT an Electric Field because: The arrows represent force and the density of lines shows the electric field strength,
which is lacking in the figure. Circle B: 20. Simple Force Potential Question A. This could be an Electne Freld. B. This is NOT an Electric Field because: The parallel lines represent equipotential lines, which are perpendicular to electric field lines. Circle A: 21. Simple Force Potential Question A.
This could be an Electric Field. B. This is NOT an Electric Field because: There is no charge represented in the figure; hence, it does not represent an electric field.
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Trial 1 shows a 1. 691 gram sample of cobalt(ii) chloride hexahydrate (mw = 237. 93). What mass would we expect to remain if all the water is heated off?
We would expect approximately 0.921 grams to remain after heating off all the water from the cobalt(II) chloride hexahydrate sample.
To calculate the expected mass remaining after heating off all the water from the cobalt(II) chloride hexahydrate sample, we need to determine the mass of water in the compound and subtract it from the initial sample mass.
The formula for cobalt(II) chloride hexahydrate is CoCl2 · 6H2O, indicating that there are 6 water molecules associated with each molecule of cobalt(II) chloride.
The molar mass of cobalt(II) chloride hexahydrate can be calculated as follows:
Molar mass = (molar mass of Co) + 2 * (molar mass of Cl) + 6 * (molar mass of H2O)
= (58.93 g/mol) + 2 * (35.45 g/mol) + 6 * (18.02 g/mol)
= 237.93 g/mol
Given that the initial sample mass is 1.691 grams, we can calculate the mass of cobalt(II) chloride hexahydrate using its molar mass:
Number of moles = mass / molar mass
= 1.691 g / 237.93 g/mol
= 0.00711 mol
Since each mole of cobalt(II) chloride hexahydrate contains 6 moles of water, the moles of water in the sample can be calculated as:
Moles of water = 6 * number of moles of cobalt(II) chloride hexahydrate
= 6 * 0.00711 mol
= 0.0427 mol
The mass of water can be calculated by multiplying the moles of water by the molar mass of water (18.02 g/mol):
Mass of water = moles of water * molar mass of water
= 0.0427 mol * 18.02 g/mol
= 0.770 g
Finally, we can calculate the expected mass remaining after heating off all the water by subtracting the mass of water from the initial sample mass:
Expected mass remaining = initial sample mass - mass of water
= 1.691 g - 0.770 g
= 0.921 g
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Calculate the force in lb, required to accelerate a mass of 7 kg at a rate of 17 m/s²?
Therefore, the force required to accelerate a mass of 7 kg at a rate of 17 m/s² is approximately 26.78 lb.
To calculate the force required to accelerate a mass of 7 kg at a rate of 17 m/s², you can use the formula F = ma, where F is the force in newtons, m is the mass in kilograms, and a is the acceleration in meters per second squared. Since the question asks for the force in lb, we will need to convert the result from newtons to pounds.
First, we can calculate the force in newtons by multiplying the mass by the acceleration: F = 7 kg x 17 m/s² = 119 N.
To convert newtons to pounds, we can use the conversion factor 1 N = 0.2248 lb. Therefore, the force required to accelerate a mass of 7 kg at a rate of 17 m/s² is:
F = 119 N x 0.2248 lb/N = 26.78 lb.
Therefore, the force required to accelerate a mass of 7 kg at a rate of 17 m/s² is approximately 26.78 lb.
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Find the total resistance of the combination of resistors
if A=150 Ω , B=730 Ω,, and C=370Ω .
A B C are side to side
Ω=
The total resistance of the combination of resistors is 1250 Ω.
To get the total resistance of a combination of resistors that are connected in a row, it is essential to follow these two steps:Add all the resistors values together to get the equivalent resistance. In this case,
AB = A + B = 150 Ω + 730 Ω = 880 Ω ABC = AB + C = 880 Ω + 370 Ω = 1250 Ω
Therefore, the total resistance of the combination of resistors is 1250 Ω.
This means that the flow of current through the resistors will face the resistance of 1250 Ω, which will limit the flow of the current to some extent.
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At t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 31.0 rad/s until a circuit breaker trips at time t = 1.50 s. From then on, it turns through an angle 433 rad as it costs to a stop at constant angular acceleration.
Part A Through what total angle did the whol turn between t = 0 and the time stopped? Express your answer in radians
θ = _____________ rad
Part B At what time did it stop? Express your answer in seconds ? t = ____________________ s
At t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 31.0 rad/s until a circuit breaker trips at time t = 1.50 s and it turns through an angle 433 rad, then the total angle with which the wheel turn between t=0 and the time stopped is θ = 227.012 rad and the time at which it stops is t= 7.79 s.
A grinding wheel has an initial angular velocity, ω₁ = 26.0 rad/s, Constant angular acceleration, α = 31.0 rad/s², Time after which the circuit breaker,
Let, the final angular velocity of the wheel be ω₂.
Final angular velocity, ω₂ = 0 rad/s
a)
We need to find the total angle through which the wheel turns between t = 0 and the time it stops.
Total angle through which the wheel turns between t = 0 and the time it stops is given by,
θ = θ₁ + θ₂
where, θ₁ = angle moved by the wheel before circuit breaker trips, θ₂ = angle moved by the wheel after circuit breaker trips
θ₁ = ω₁t + 1/2 αt²
where, ω₁ = initial angular velocity, t = time taken for circuit breaker to trip, α = angular acceleration
θ₁ = 26.0(1.50) + 1/2(31.0)(1.50)²= 113.625 rad
θ₂ = ω² - ω²/2α
where,ω = initial angular velocity = 26.0 rad/s
ω₂ = final angular velocity = 0 rad/s
α = angular acceleration= 31.0 rad/s²
θ₂ = (26.0)²/2(31.0)= 114.387 rad
Total angle through which the wheel turns between t = 0 and the time it stops,
θ = θ₁ + θ₂= 113.625 + 114.387= 227.012 rad
Therefore, the total angle through which the wheel turns between t = 0 and the time it stops is 227.012 rad.
b) We need to find the time at which it stops.
Using the relation,
θ = ω₁t + 1/2 αt²θ - ω₁t = 1/2 αt²t = √2(θ - ω₁t)/α
At t = 0, the wheel has an angular velocity, ω₁ = 26.0 rad/s
So,The time it stops, t = √2(θ - ω₁t)/α= √2(433 - 26.0(1.50))/31.0= 7.79 s
Therefore, the wheel stops at t = 7.79 s.
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Build a circuit that has an adjustable power supply that adjusts the output voltage from 0 volts to 15 volts, and also has a fixed 8 volt power output. And also the supply will power a circuit containing a transistor or op amp
It is also necessary to make a description of the operation of the circuit
A circuit that can provide an adjustable power supply that can adjust the output voltage from 0 volts to 15 volts and also provide a fixed 8 volt power output, as well as power a circuit containing a transistor or op amp can be built using the following components and operation steps:
Components needed:
One transformer
One bridge rectifier
One 4700 uF capacitor
Two 1000 uF capacitors
One 15k potentiometer
One 12V Zener diode
One NPN transistor
One 10k resistor
Two 1k resistors
Operation description:
1. Begin by connecting the transformer's primary winding to the mains and its secondary winding to the rectifier circuit. The transformer should have a 12-0-12 volts, 1A secondary winding.
2. The bridge rectifier is connected to the secondary winding, which is composed of four 1N4007 diodes, with two of them mounted in one direction, while the other two are mounted in the opposite direction.
3. A 4700 uF capacitor is connected across the bridge rectifier's output to remove the ripple component of the rectified signal.
4. The 12V Zener diode is connected in parallel with the two 1000 uF capacitors, which are connected in series, with one side of each capacitor connected to one end of the potentiometer. The other ends of both capacitors are joined together and connected to the 0V terminal.
5. The potentiometer's center wiper is linked to the output, while one end is linked to the input.
6. A 10k resistor is connected between the input and the base of the transistor, with the collector of the transistor connected to the output and the emitter linked to the 0V terminal.
7. Finally, two 1k resistors are used to bias the op amp circuit, with one resistor connected between the input and the op amp's positive input and the other resistor connected between the negative input and the 0V terminal.
In this configuration, the output voltage can be changed by moving the potentiometer's wiper to any point between the input and 15 volts. The 8 volt output is fixed and is located between the input and the potentiometer's 0 volt output. The op amp circuit is also biased by two 1k resistors.
Thus the required connection is set up.
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