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2 years ago

Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.

Physics

1 answer:

mote1985 [20]2 years ago

7 0

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is $W_c = P_o V_o ln (R_v)$

Explanation:

Generally smallest workdone done by a gas is mathematically represented as

$dW = PdV$

Generally for an isothermal process

$PV = nRT = constant$

=> $P = \frac{nRT}{V}$

Generally the total workdone is mathematically represented as

$W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV$

=> $W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV$

=> $nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.$

=> $W_c = nRT [ln(V_f) - ln(V_o)]$

=> $W_c = nRT ln \frac{V_f}{V_o}$

From the question $\frac{V_f}{V_o } = R_v$

=> $W_c = P Vln (R_v)$

at initial state

$W_c = P_o V_o ln (R_v)$

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A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

Similarly, the distance y the object has traveled is calculated as follows:

$\displaystyle y=\frac{g\cdot t^2}{2}$

If we know the height h from which the object was dropped, we can solve the above equation for t:

$\displaystyle t=\sqrt{\frac{2\cdot y}{g}}$

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

$\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec$

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

$t_2=2.56 - 2 = 0.56\ sec$

Therefore, it has traveled down a distance:

$\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m$

Thus, the height of the pen is:

$h_p = 32 - 1.54\Rightarrow h_p=30.46\ m$

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2 years ago

A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

$T - mg = ma$

here we know that maximum possible acceleration of so that string will not break is given as

$T = 54 N$

now we have

$54 - (5 \times 9.8) = 5 a$

$a = 1 m/s^2$

now for such acceleration we can use kinematics

$d = \frac{1}{2}at^2$

$2 = \frac{1}{2}(1) t^2$

t = 2 s

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2 years ago

If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?

frutty [35]

Answer:

So length of pendulum is 143.129 m

Explanation:

We have given period of simple pendulum is 2 sec

We have to find the length of simple pendulum

Let the length of pendulum is l

Acceleration due to gravity $g=9.8m/sec^2$ is

Time period is given by $T=2\pi \sqrt{\frac{l}{g}}$

So $24=2\times 3.14\times \sqrt{\frac{l}{9.8}}$

$\sqrt{\frac{l}{9.8}}=3.821$

Squaring both side

${\frac{l}{9.8}}=14.60$

l =143.129 m

So length of pendulum is 143.129 m

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2 years ago

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Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b

Ainat [17]

Answer: the pair of sunglasses

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A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

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2 years ago

A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc

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Change in time = d(t)
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t1 = 0.00 s
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Average acceleration = d(v) / d(t)
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2 years ago

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