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Svetradugi [14.3K]
3 years ago
15

When one bumper car hits another, how does the reaction force affect the first car?

Physics
2 answers:
Ksenya-84 [330]3 years ago
8 0

<u>Answer:</u> The correct answer is Option a.

<u>Explanation:</u>

We are given that two bumper cars hits each other and we need to find the affect caused on first car by the reaction force.

According to the Third Law of Newton,

For every action there is an equal and opposite reaction.

This law actually means that in every interaction, two forces are acting on both the interacting objects.

Hence, the the car collided will move opposite to the direction of action force. Therefore, the correct answer is Option a.

Colt1911 [192]3 years ago
3 0
A) The car moves in the direction opposite of the action force.

By Newton's 3rd Law, for every action, there is an equal and opposite reaction.
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What connects the upper motor neurons to lower motor neurons?
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The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

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Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

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Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?
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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
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Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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