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SOVA2 [1]
3 years ago
5

The video shows an animated billiards experiment in which a cue ball strikes a glued-in-place eight-ball. which of the following

explains why the momentum of the eight-ball is conserved?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0
Suppose object A<span> is a cue ball and object </span>B<span> is an eight ball on a pool table. If the cue ball strikes the eight ball, the cue ball exerts a force on the eight ball that sends it rolling toward the pocket. At the same time, the eight ball exerts an equal and opposite force on the cue ball that brings it to a stop. Note that both the cue ball and the eight ball each experience a change in momentum. However, the sum of the momentum of the cue ball and the momentum of the eight ball remains constant throughout.</span>
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Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2
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The particles can undergo small oscillations around x₂.

The given parameters;

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3 0
2 years ago
In grinding a steel knife blade (specific heat = 0.11 cal/g-c),the metal can get as hot as 400C. If the blade has a mass of 80g,
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33 g.

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Assuming no heat transfer can be possible except for heat exchange between water and steel, we can say that the heat lost by the knife, must be equal to the heat gained by the water.

As we have a limit for the maximum temperature of both elements (once reached a final thermal equilibrium), of 100ºC, which means that the maximum allowable change in temperature will be of 300º C for the knife, and of 80º C for the water.

Empirically , it has been showed that for a heat exchange process using only conduction, the heat needed to raise the temperature of a body, is proportional to the mass, being the proportionality constant a factor that depends on the material, called specific heat.

So, we can write the following equation:

cs*mk*Δtk = cw*mw*Δtw

Replacing by the givens of the question, we have:

0.11 cal/gºC * 80 g * 300ºC = 1 cal/gºC*mw*80ºC

Solving for mw = 2,640 cal / 80 cal/g =33 g.

5 0
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