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course phy201
`q001. Describe briefly what you did in lab today and show you data.We balanced dominoes on both sides of a ramp to get different variation of data we balanced a single domino and 2 on the other side. Then for another we placed 2 dominoes on both sides to get different points where the system was in balance
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What did you learn from doing this brief experiment?
The location of the domino in comparison to the balancing pt determines how much weight it can balance on the other side of the strap
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Pick four of your trials, choosing the most varied among the situations you set up.
Assume the weight of a domino is 1 dominoWeight. If you really want to use a weight in Newtons, you can use .16 Newtons, but dominoWeight is fine.
For your first chosen trial, what was the net torque exerted by the dominoes? Be sure to give the details of your calculations.
-M1gxD1
1dw(9.8m/s^2)x21cm
Dm1=-205.8
-m2gxd2
Dw(9.8m/s^2)x22cm
Dm2=-215.6
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What was the net torque for this trial, as a percent of the sum of the magnitudes (i.e., absolute values) of the torques?
-205.8+215.6=0
9.8Nm
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You won't end up with N * m as your unit. Your calculation was dw * m/s^2 * cm, so your units are dw * m * cm / s^2.
You wouldn't actually multiply the domino weight by 9.8 m/s^2 in the first place; the domino weight is already a force, not a mass. Were you to calculate in terms of domino mass, you would have to multiply by the acceleration of gravity to get domino weight.
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Report the net torque and the net torque as a sum of the magnitudes of the torques for the second chosen trial. You have already explained the details of how you do the calculations, and don't need to include the details here or for the next two trials.
2dm(9.8ms2)(9)cm
-176.4
1dm(9.8m/s^2)
-176.4
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This calculation would require a distance, which I assume was 18 cm.
Note that your data were to have been reported in answer to the first question.
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Net torque=0
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Report the net torque and the net torque as a sum of the magnitudes of the torques for the third chosen trial.
1dwx9.8(20)
19.6
1dw(9.8)(20)
19.6
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Report the net torque and the net torque as a sum of the magnitudes of the torques for the fourth chosen trial.
2dw(9.8)(6)
1.18
1dw(9.8)(20.5)
2.0
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`q002. A projectile has initial upward vertical velocity 5 meters / second. It is released at a height of 2 meters above the ground. Its acceleration is -9.8 m/s^2, as is the case for all ideal projectiles. How long does it take to reach the ground?
A=-9.8m/s2
V0=5m/s
Ds=2m
Vf=3.76
Vave=4.38
Dt=.46s
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You haven't indicated how you calculated this, but an ideal projectile with a nonzero initial velocity in the vertical direction, which lands below its initial point, will end up with a final velocity greater than its initial.
Your calculation used
A=-9.8m/s2
V0=5m/s
Ds=2m
which indicates a displacement in the direction opposite the acceleration; in this case that would entail an upward displacement, so the projectile would land 2 meters above the initial position. Your calculation of vf would be correct for this situation, but not for the given situation in which the projectile lands 2 meters below its initial position.
The final velocity will in any case be in the direction opposite the initial velocity, so the average velocity will be an average of two velocities of opposite sign.
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`q003. A projectile is released from a height of 4 meters and requires 1.2 seconds to reach the ground. Its initial horizontal velocity is 12 meters / second. How far does it travel in the horizontal direction before hitting the ground?
V0=12m/s
Dt=1.2
Ds=4m
12m/sx1.2s
Travels 14.4m in horizontal direction before hitting the ground
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`q004. What would be the initial vertical velocity of a projectile which, when released from a height of 4 meters, requires 1.2 seconds to reach the ground?
The angle is about 70degress
Vy=12m/scostheta
11.3m/s
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The question asks only about the vertical motion. There is no information related to 70 degrees. It's not clear what you used to find that angle.
This is just a uniform-acceleration problem, where you know `ds, `dt and a.
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`q005. Just before striking the ground a projectile has a vertical velocity of 10 m/s, downward, and a horizontal velocity of 15 m/s. What are the magnitude and direction of its velocity at that instant, with direction measured on an x-y coordinate system with a vertical y axis?
A2+b2=c2
C=18
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That's a reasonable result for the magnitude of the velocity, but doesn't mention the direction.
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If the mass of the projectile is 5 kg, what is its kinetic energy at the instant it strikes the ground?
Ke=1/2mv^2
KE=62.5 J
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It's not clear what you used for mass and velocity, but for a projectile of mass 5 kg moving at 18 m/s the KE would be a lot more than 62.5 Joules.
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What is its kinetic energy in the x direction (i.e., its kinetic energy calculated just on the basis of its horizontal velocity?)
1/2x5x15^2
562.5 J
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What is its kinetic energy in the y direction (i.e., its kinetic energy calculated just on the basis of its vertical velocity?)
Ke=1/2x5x-10^2
Ke=250J
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What is the sum of its x and y kinetic energies?
X+y
Ke=812.5J
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This appears to be correct.
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You should recalculate the KE at the instant of striking the ground, where the velocity as you have calculated it is 18 m/s.
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`q006. The washer I tossed into the trash can appears to have had an initial velocity of about 5 meters / second and left my hand at an angle of about 20 degrees above horizontal. The trash can was 5 meters away, the washer was released at a height of 170 cm and entered the trash can at a height of 80 cm.
Are these quantities consistent (remember that our original estimate of 80 cm/s at 30 degrees was consistent with hitting a trash can about a foot away)? Specifically, would the given initial velocity, angle, height of release, and height of the trash can result in reaching the 80 cm height after traveling 5 meters? If not, at what distance would this 'shot' reach the 80 cm height?
Vx=50cmcos160
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Do you mean 50 cm/s?
Note that 50 cm is half a meter; 5 meters is a lot more than 50 cm.
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-47cm/s
Vy=50cmsin160
17.1cm/s
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Suggested procedure (if your solution above is complete you don't have to answer these individual questions; if not, you should):
It will reach the 80 cm height after traveling 5m
Find the initial vertical and horizontal velocities.
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Figure out what other two vertical quantities are known from the given information.
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Analyze the vertical motion to find the time required to get to the 80 cm height.
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Apply this time to the horizontal motion.
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`q007. Give a brief synopsis of what you learned by using the PHeT program.
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Check my notes. You're on the right track but are missing some details.
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