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3 years ago

6

An ideal gas in a balloon is kept in thermal equilibrium with its constant-temperature surroundings. How much work is done if th

e outside pressure is slowly reduced, allowing the balloon to expand to 50 times its original size

1 answer:

andreev551 [17]3 years ago

5 0

Answer:

w = 252.32 N

Explanation:

given data

balloon expand = 50 times its original size

we consider here initially pressure and volume

pressure = 645 pa

volume = 0.10 m³

solution

as in isothermala process ideal gas

PV = mRT

P = $\frac{mRT}{v}$

P = $\frac{c}{v}$

here c is constant

so work done is express as

$w = c \int\limits^{V2}_{V1} {\frac{dv}{v}}$

w = $c \times ln( \frac{v2}{v1})$

and we know c = p1 × v1

so

w = p1 × v1 × $ln (\frac{50v1}{v1} )$

w = 645 × 0.1 × ln(50)

w = 252.32 N

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A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

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Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

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Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

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Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

$Thickness=\frac {\lambda}{4\times n}$

Where, n is the refractive index of the film

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {136.8\ \mu\ m}{4\times 19}$

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

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Answer:

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Explanation:

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Answer:

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