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miss Akunina [59]
3 years ago
14

4. Both trains speed up all the time.

Physics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

The velocity with which the ball strikes the ground = -5.7 m/s

The velocity with which the ball leaves the ground = 4.3 m/s

Explanation:

Seems like there's parts of a couple questions. I'll focus on the tennis ball part which seems to be a full question.

To find the velocity with which the tennis ball hits the ground, we only need to worry about what happens up to that point. We can ignore the rebound for this part. Given:

d = -1.85

a = -9.8

vi = 0

vf = ?

v_f^2 = v_i^2 + 2ad\\v_f^2=0+2(-9.8)(-1.65)\\v_f^2=32.34\\v_f=-5.7

*Keep in mind that the square root gives us two answers, a positve and a negative one. We use the negative one here because the final speed is downwards and the question says down is negative.

To find the velocity with which the ball leaves the ground, we only need to worry about what happens after that point. We can ignore the drop for this part. Given:

d = 0.947

a = -9.8

vf = 0

vi = ?

v_f^2 = v_i^2 + 2ad\\0=v_i^2+2(-9.8)(0.947)\\v_i^2=18.56\\v_i=4.3

*Here we use the positive speed after the square root because we know the ball is going up, which the question designates as the positive direction.

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4 0
2 years ago
A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe
Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

\Delta U=-5 J is the loss in potential energy of the system

\Delta K=+6 J is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

E=K+U

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

\Delta E= \Delta K + \Delta U

In this problem:

\Delta K=+6 J

\Delta U=-5 J

So, the change in mechanical energy is:

\Delta E=+6+(-5)=+1 J

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

W=\Delta E=+1 J

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F-spondin.

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Explanation:

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miss Akunina [59]
1. the amount of electric charge that they carried
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