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Whitepunk [10]
3 years ago
12

For an impulse-momentum experiment, students collect data on a collision between blocks A and B . A force probe attached to bloc

k A is used to measure the force exerted by block A on block B as a function of time during the collision. The data collected shows a positive force that is used to determine the impulse on block B . The force sensor is removed from block A and attached to block B , and the experiment is run again. Which of the following correctly describes how this affects the data and if the impulse on block B can still be determined?
A. The data can only be used to determine the impulse on block A, not block B.
B. The data can only be used to determine the impulse on block B, not block A.
C. The data cannot be used to determine the impulse on block A or block B.
D. The data can be used to determine the impulse on block B and block A.
E. The data can be used to determine the force on block B.
Physics
1 answer:
svp [43]3 years ago
3 0

Answer:

the answer the correct one is D

Explanation:

The relationship between momentum and moment is

         I = Δp

         I = F t

This is for one of the body, that is, the force and the change in speed is for the same body.

In this case, in the first part, the sensor is placed in body A, therefore the force of this body can be measured and as time is also measured, let's calculate the impulse of body A.

In the second part of the experiment, the probe is placed in body B and the experiment is repeated, so we can also calculate the momentum of body b.

consequently in this experiment the momentum of body a and b can be measured.

to check the answer the correct one is D

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4 0
3 years ago
1. Is the collision between the ball and the pendulum elastic or inelastic? Justify your answer by calculating the kinetic energ
Nitella [24]

Answer:

So energy is not conserved and inelastic shock

Explanation:

In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.

Let's find the kinetic energy just before the crash

         K₀ = ½ m vₓ₀²

After the crash we can use the law of conservation of energy

Starting point. Right after the crash, before starting to climb

        Em₀ = K = ½ (m + M) v₂²

Final point. At the maximum height of the pendulum

       Em_{f} = U = (m + M) g h

Where m is the mass of the bullet and M is the mass of the pendulum

        Em₀ = Em_{f}

        ½ (m + M) v₂² = (m + M) g h

        v₂ = √ 2g h

Now we can calculate the final kinetic energy

       K_{f} = ½ (m + M) v₂²²

       K_{f} = ½ (m + M) (2gh)

The relationship between these two kinetic energies is

      K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)

      K₀ / K_{f} = m / (m + M)  vₓ₀² / 2 g h

We can see that in this relationship the Ko> Kf

So energy is not conserved and inelastic shock

5 0
4 years ago
5. Which of the following is velocity? *
ivann1987 [24]
A. 20m/s because the unit for velocity is m/s
3 0
3 years ago
A brass cube, 10 cm on a side, is raised in temperature
lana [24]

Answer:

\delta L\%=22.5\%

Explanation:

<em>Assuming that the thermal expansion of brass cube occurs isotropically (i.e. equal in all the directions).</em>

Given:

  • length of the cube, l=10\ cm=0.1\ m
  • change in temperature of the cube, \Deta T=200\^{\circ}C
  • coefficient of volume expansion, \beta=57\times 10^{-6}\ ^{\circ}C^{-1}

<u>Hence volume of the cube:</u>

V=10^{-3}\ m^3

Now the volume of the cube after expansion:

\delta V=V.\beta.\Delta T

\delta V=10^{-3}\times 57\times 10^{-6}\times 200

\delta V=1.14\times 10^{-5}\ m^3

Therefore,

\delta L=0.0225\ m

<u>Now the percentage change in the edges of the cube:</u>

\delta L\%=\frac{\delta L}{L} \times 100

\delta L\%=\frac{0.0225}{0.1} \times 100

\delta L\%=22.5\%

5 0
3 years ago
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