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aalyn [17]
3 years ago
8

A certain amount of a monatomic gas is maintained at constant volume as it is cooled by 50K. This feat is accomplished by removi

ng 400 J of energy from the gas. How much work is done by the gas?
a) 400 J
b) -400 J
c) zero
d) none of the above
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

The correct answer is option a. 400J

Explanation:

Hello!

Let's solve this!

To know if the work is positive or negative we have to know the following:

The work is positive when the gas performs (expansion)

The work is negative when done outside on the gas (compression)

In this case the work is done by the gas, so it is positive.

The correct answer is option a. 400J

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Explanation:

(1) Every object moves in a straight line unless acted upon by a force. (2) The acceleration of an object is directly proportional to the net force exerted and inversely proportional to the object's mass. (3) For every action, there is an equal and opposite reaction.

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A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What
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Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let p_{1} = momentum of the 1st ball

p_{2} = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}

or

(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}

Since we are dealing with identical balls, all the m terms cancel out so we are left with

(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} +  (v_{2})_{f}\cos \theta_{2}

Putting in the numbers, we get

1.33 = (0.750) \cos(33.30)  + (v_{2})_{f} \cos \theta_{2}

=  > (v_{2})_{f} \cos \theta_{2} = 0.703

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}

or

=  > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30)  = 0.412

Taking the ratio of the sine equation to the cosine equation, we get

\frac{ \sin \theta _{2}}{ \cos \theta_{2} }  =  \tan \theta_{2}  =  \frac{0.412}{0.703}  = 0.586

or

\theta_{2}  =  { \tan}^{ - 1} (0.586) = 30.4

Solving now for (v_{2})_{f},

(v_{2})_{f}  =  \frac{0.412}{ \sin(30.4) }  = 0.815 \:  \frac{m}{s}

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Answer:

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