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amid [387]
3 years ago
7

A perfectly still 0.01 kg marble is hit with a force of 50 N by a pencil. According to Newton's third law, which describes the f

orce the marble exerts on the pencil? A. The marble exerts a force of 0.05 N on the pencil. B. The marble exerts a force of 0.5 N on the pencil. C. The marble exerts a force of 5 N on the pencil. D. The marble exerts a force of 50 N on the pencil.
Physics
1 answer:
WARRIOR [948]3 years ago
5 0

During the time that the pencil is exerting 50N of force on the marble, the marble is exerting 50N of force in exactly the opposite direction on the pencil. <em>(D)  </em>The mass of the marble doesn't matter, and it doesn't even matter whether the marble is moving or perfectly still.

As soon as the pencil stops exerting any force on the marble, the marble immediately stops exerting any force on the pencil.

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Tresset [83]
For this case we first think that the skateboard and the child are one body.
 We have then:
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 2 = skateboard + boy
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 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
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 answer:
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3 years ago
Which force requires contact?
Minchanka [31]

Answer:

A

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

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Here Temperature of soup is constantly decreasing

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efficiency is given by

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dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

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