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3 years ago

A student is sitting at rest in a chair. How does the force that the student exerts

Physics

1 answer:

jonny [76]3 years ago

3 0

Answer:

the same magnitude but the opposite direction

Explanation:

Newton's third law of motion states that there is always an equal and opposite reaction to every action. This means that the amount of force exerted upon an object is equal to the amount of force the object exerts but in an opposite direction.

This is the case in this scenario where a student sits at rest in a chair. The student is supplying the action force being exerted on the chair. According to the third law of Newton, the chair will exert the same size of force back in the student but in an opposite direction.

Hence, the force the chair exerts on the students compare with that of the student in the sense that they are the same magnitude (size) but the opposite directions.

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A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m

Whitepunk [10]

Answer:

<h2>The pin's final velocity is 5m/s</h2>

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

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In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b

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The total average velocity $v=+1.41 m/s$ (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
$v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}$
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion ( $S_1=6.30 km=6300 m$ , due west) and of the displacement in the second part of the motion ( $S_2$ , due east).
-The total time taken t is the time taken for the first part of the motion, $t_1$ , and the time taken for the second part of the motion, $t_2$ . $t_1$ can be found by using the average velocity and the displacement of the first part:
$t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s$

$t_2$ , instead, can be written as $\frac{S_2}{v_2}$ , where $v_2=-0.630 m/s$ is the average velocity of the second part of the motion (with a negative sign, since it is due east).

Therefore, we can rewrite the initial equation as:
$v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }$
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$S_2=-844 m=-0.844 km$

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3 years ago