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4 years ago

Question 2 An ideal gas in a piston/cylinder device is compressed at constant temperature. The entropy of the ideal gas will:

Physics

1 answer:

Flura [38]4 years ago

8 0

To solve the problem it is necessary to refer to the definition of entropy.

Entropy is defined aso

$\Delta S = \frac{\Delta Q}{T}$

Where,

$\Delta Q =$ Heat exchange

T = Temperature

Since the heat exchange is conserved and it is an isothermal process where the temperature remains constant the change in entropy remains the same, ie $\Delta S = 0$ (Reamins same)

Therefore the correct answer is C.

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Mark and Paul are in a race. Mark is 20 meters from the finish line and running at a constant 3.5 m/s. Paul is 5 meters behind h

USPshnik [31]

Answer:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

Explanation:

For this case we have an illustration for the problem on the figure attached.

And we can solve this problem analyzing each one of the runner. Let's begin with Mark

Mark

For this case we know that $V_M = 3.5 m/s$ and th velocity is constant. The distance from Mark and the finish line is $D_M = 20 m$

Since the velocity is constant we can create the following relation:

$D_M = V_M t_M$

And solving for $t_M$ we got:

$t_m = \frac{D_M}{V_M}= \frac{20m}{3.5 m/s}= 5.71 s$

So then Mark will nd the race after 5.71 seconds

Paul

We know that the initial velocity for Paul is given $V_{iP}= 2.7 m/s$ we also know that the total distance between Paul and the finish line is 25 m and we want to find the acceleration that Paul needs to apply in order to tie the race, and Paul have 5.71 sconds in order to reach the finish line.

We can use this formula in order to find the acceleration (because we assume that the acceleration is constant) that he needs to apply:

$x_f = x_i + v_i t + \frac{1}{2} a t^2$

And since $\Delta x = x_f - x_i$ we have this:

$\Delta x= v_i t + \frac{1}{2} a t^2$

And if we replace we have this:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

And the final velocity for Paul using this acceleration would be:

$V_{fP}= V_{iP}+ a_P t = 2.7m/s + 0.588 m/s^2 (5.71s)= 6.057 m/s$

3 0

4 years ago

Physics Help Please!!! All multiple choice!

Natasha_Volkova [10]

1) True.
Uniform circular motion means that the angular velocity $\omega$ is constant, and since the centripetal acceleration is given by
$a=\omega ^2 r$
if $\omega$ is constant, then a is constant.

2) The correct answer is
<span>a.) the angular velocity and linear velocities are the same
The three knots are in fact on a circle, therefore the angular velocities are the same (because they cover the same angle in the same time) and the linear velocities are the same as well, because it is given by
</span> $v=r \omega$
and since both r(distance from the center) and $\omega$ are the same, v is also the same.

3) <span>b.) toward the center
In fact, an object can move on a circular path only if there is a force pushing towards the center. This force is called "centripetal force".</span>

3 0

3 years ago

what is the property of an object due to its mass by which it resists any change in its position unless overcome by force ?

UkoKoshka [18]

Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

hope this helps

5 0

3 years ago

Which sentence in the passage can be used to conclude that Eris is a dwarf planet and not a planet?