For what time interval does the exam last as measured by the students?
The issue is how do we find the ||\gamma||. In this question, both the students and the professor are moving with velocities relative to the earth. The velocity of the professor relative to the students was not given directly. We need to find that velocity. Set up the reference frame shown below:
The earth is the rest frame set as O and both the professor and the students are traveling relative to the earth. Set the students' frame as O', which is moving with v=0.7c in +x relative to the earth. Now we need to find the velocity (u') of the professor traveling relative to the students. Use Lorentz transformation on velocity (shown in the diagram), we can find u'.
With u' found, we can then take the students' frame and make observation of the professor's time:
Here with u' known, we can use time dilation to find the students' observation of the professor's time. The professor holds the clock and has the proper time. We then have: $$\Delta t'=\gamma \Delta t_p$$
]]>A spaceship traveling at 0.540c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.880c relative to the mother ship. As measured on the Earth, the spaceship is 0.280 ly from the Earth when the landing craft is launched.
(a) What speed do the Earth-based observers measure for the approaching landing craft?
(b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the mother ship?
(c) What travel time is required for the landing craft to reach the Earth as measured by the mother ship?
(d*) What travel time is required for the landing craft to reach the Earth as measured by Earth-based observers?
(e*) What travel time is required for the landing craft to reach the Earth as measured by those on the landing craft?
Answers
For relativity problems, the first step is to set up coordinate systems and clearly define all the velocities and the reference frames the velocity is relative to.
(a) In this part, set the earth being the rest frame (O frame), the mother ship as O' frame. Then v=0.540c and the landing craft relative to the mother ship is u'=0.880c. To find the landing craft's velocity relative to the earth, we can use the Lorentz transformation on velocity: $$u=\frac{u'+v}{1+vu'/c^2}$$
(b) From the mother ship's point of view, the earth is moving toward's mother ship at v=-0.540c, therefore, we have $$\Delta {x'}=\frac{\Delta x}{\gamma}$$
Here ||\Delta x|| is the length measured in the earth frame.
(c) From the mother ship, the distance between the earth and the landing craft is ||\Delta{x'}||, the landing craft is moving in +x direction with u'=0.880c relative to the mother ship and the earth is moving towards the -x direction with u'_{e}=-0.540c. See the diagram below:
The time for the landing craft to meet the earth calculated by the observer on the mother ship would be $$\Delta{t'}=\frac{\Delta{x'}}{u'-u'_e} $$
(d) From the earth based observer, the landing craft is traveling at a velocity u found in part (a). The distance between the landing craft and the earth is given as 0.28 ly when the landing craft is released. Therefore, the time takes the landing craft to meet the earth calculated in earth frame is $$\Delta t=\frac{\Delta x}{u}$$
(e) For those on the landing craft, the earth's velocity is ||v=-u||. Here u is the velocity found in part b. Now we need to find the distance between the earth and the landing craft as observed by the landing craft.
We can use this velocity to calculate ||\gamma|| for the observer on the landing craft to observe earth events. Then we can calculate the distance with ||\Delta{x''}=\frac{\Delta x}{\gamma}||. Then the time observed on the landing craft is $$\Delta{t''}=\frac{\Delta{x''}}{v}$$
]]>See the diagram for the set up.
We can use Bernoulli equation to solve for the v. Here we can assume that the cross section area of the tube is much smaller than that of the beaker so we can treat the velocity that the water level drops in the beaker to be zero. Setting y=0 at the siphon's outlet, we have $$P_0+\rho gh=P_0+\frac 12\rho v^2$$
Plug in numbers and we can get v.
To find out the highest height the siphon can go, the question becomes equivalent to the following diagram. Basically, the liquid will evaporate into vapor gas and create a pressure — that is the lowest pressure it can get to. Lower than that, the liquid will boil to maintain such a pressure. For example, the liquid would instantaneously boil and evaporate in vacuum at room temperature. So in this question, the P at the top of the siphon or the test tube shown below would be equal to the vapor pressure.
With ||P_A=P_B=P_0||, we have ||P_A=P_0=\rho gy+P||.
Plug in the numbers and we can find y here.
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]]>Consider a hydraulic brake system as shown below. The area of the piston in the master cylinder connected to the foot paddel is 1.8 cm2. The area of the piston connected to the brake shoe is 6.4 cm2. The kinetic frictional coefficient between shoe and wheel drum is 0.50. If the wheel's radius is 34 cm, find the frictional torque about the axle when a force of 47 N is exerted on the brake pedal.
Answer solution:
For the hydraulic brake system, first we need to find the force the brake shoe is pushing the wheel drum. This can be done using the hydraulic pressure and force relation. Here P is same in the hydraulic system and the pressure force exerted on the cylinders are ||F=PA||, where A is the area of the cylinder. See the figure below for illustration of the forces and torque on the brake system.
Since pressure in the hydraulic system is the same, we have $$\frac{F_p}{A_1}=\frac{N}{A_2}$$
This will allow you to find the normal force the shoe pushes on the wheel drum.
Then this normal force will create friction ||f=\mu mg||, which then creates a torque ||\tau=fR||.
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Consider the circuit shown in the diagram, find (A) the current in the 2.00 Ω resistor, (B) the current in the 6.00 Ω resistor, and (C) the potential difference between points a and b ). Identify the current directions as to the right or to the left.
Doing a little rearrangement, we have the following circuit diagram:
At node b, we have ||I_1=I_3+I_2||.
For the two loops, we have $$12-4I_1-2I_3=0$$ and $$8+2I_3-6I_2=0$$
Solve for for the ||I||'s.
Then we can find ||V_{ab}=2I_3||.
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What are the signs of Q , W_{done}_{ on}, and ΔE_{int} that are associated with each process, and for the process as a whole?
Q | W_{ongas} | ΔU (ΔE_{int}) | |
A to B | |||
B to C | |||
C to A | |||
Complete |
From A to B: The gas is heated to increase its volume and pressure, so Q is positive and ||W=-P\Delta V|| is negative. Temperature increases so ||\Delta U|| is positive.
From B to C: Work is zero since ||\Delta V=0||. Need to cool it to reduce pressure and the temperature also drops on the PV diagram (from an isothermal line with higher temperature to an isothermal line with lower temperature). So Q is negative and ||\Delta U|| is negative.
From C to A: The gas is cooled to decrease its volume, so Q is negative and ||W=-P\Delta V|| is positive. Temperature also decreases so ||\Delta U|| is negative.
For the whole cycle considering from A to B to C and back to A. The total work on the gas is negative: the work done in AB is negative and the work done in CA is positive. But the magnitude of the work in AB is larger than that is CA, so the total work to gas is negative (another way to think about this is that the gas is doing positive work to the environment).
For a complete cycle, the internal energy doesn't change (same temperature), therefore, ||\Delta U=0||. Since ||\Delta U=Q+W||, Q must be larger than zero because W is negative.
——
Q | W_{ongas} | ΔU (ΔE_{int}) | |
A to B | + | - | + |
B to C | - | 0 | - |
C to A | - | + | - |
Complete | + | - | 0 |
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(a) Determine the difference h in the heights of the two liquid surfaces.
Answer: To show more clearly how to set up the locations for calculating and comparing pressures, please reference to the following figure:
In case B, the pressures at location 1 and 2 are equal ||P_1=P_2|| and therefore $$P_0+\rho_wg(L-h)=P_0+\rho_ogL$$
In case C, the fast traveling air above the left side of the U-tube will create smaller pressure. The right side of the U-tube will still have atmosphere pressure (P_{0}) and will push the liquid up to the left side. Again, the pressures at location 1 and 2 are always equal: ||P_1=P_2||. So we have $$P+\rho_wgL=P_0+\rho_ogL.$$ Therefore $$P=P_0-(\rho_w-\rho_o)gL$$
To find the speed needed for creating the smaller pressure, we can set up the Bernoulli along the dashed line shown in the figure below, which gives $$P+\frac12\rho_{air}v^2=P_0$$
Here the vertical position (y) is identical on both sides of the equation so it is canceled. The speed of air on the right hand side is zero, so only have the ||P_0|| term.
Then we have $$\frac12\rho_{air}v^2=P_0-P=(\rho_w-\rho_o)gL.$$
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I hope that my non-calculated answers are correctly explained – be sure to look into definitions just in case.
]]>http://tutor.leiacademy.org/qa/index.php/91/901-slide-show-slide-13-15
Need to calculate the torque and then use |||tau=I\alpha||. to find ||\alpha||.
]]>This is from the lecture slides. I’m not really sure how to solve this one. Could someone give me a detailed explanation?
The linear speed depends on the distance between the point of interest and the rotational axis. And you will have ||v=\omega R||, where R is the distance between the point of interest and the rotational axis.
]]>Two stickers A and B are fixed to different locations on the disk as shown.
Compare the angular speed of the two stickers A and B.
This problem is from one of the lecture slides. Can someone give me a good explanation of why the angular speeds of A and B are the same, but why the linear speed of B is greater than A?
]]>So in the first two cases, before and after collision, the total angular momentum is the same. But after collision the moment of inertia increases, therefore, the angular speed would be smaller ||I_i\omega_i=I_f\omega_f||.
In the third case, the path of the block doesn’t go through the rotational axis and the block has a momentum in the clockwise direction opposite to the disk’s angular momentum. Therefore after the collision the angular momentum of the disk and the block together will be smaller than the angular momentum of the disk only before collision. The total angular momentum will still be the same before and after collision. ||L_{Di}+L_{Bi}=L_{(D+B)f}||.
]]>(b) momentum is always conserved in a collision as long as there is no external forces applied on the system.
(c) without frictional forces and external forces, the total mechanical energy is conserved.
(d) also conserved since the earth is also part of the system so no external forces. The earth will move towards the car with a small velocity in the opposite direction.
(e) there is an external force of friction applied on both cars so the total ME is not conserved for the two cars.
(f) there is an external force of friction so momentum is not conserved.
]]>Once have torque, you can use ||\tau_{net}=I\alpha|| to get the angular acceleration.
Here, the moment of inertia of the wheels will be approximated with a ring — ||I=mR^2||.
]]>(a) Is mechanical energy of the system conserved?
(b) Is momentum of this system conserved?
Next, consider only the process of the boxcar gaining speed as it rolls down the hump. Consider the boxcar and the Earth as a system.
(c) Is mechanical energy of this system conserved
(d) Is momentum of this system conserved?
Finally, consider the two cars as a system as the boxcar is slowing down in the coupling process.
(e) Is mechanical energy of this system conserved?
(f) Is momentum of this system conserved?
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