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2 years ago

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A sample of an ideal gas at Pi is initially confined to one chamber of the apparatus represented above, and the other chamber is

initially evacuated. The valve connecting the chambers is opened, and the gas expands at constant temperature to fill both chambers. What best describes ΔS for the process?

1 answer:

stiv31 [10]2 years ago

7 0

If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature does not change.

When the valve of the chamber opens, the ideal gas expands and since, it is free to move its entropy increases. Entropy is a measure of uncertainty or randomness. Thus, ΔS becomes positive.

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A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0

2 years ago

The gravity tractor is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to impact the Earth.

Talja [164]

Answer:

$F_g=461lb_f$

Explanation:

First calculate the mass of the asteroid. To do so, you need to find the volume and know the density of iron.

If r = d/2 = 645ft, then:

$V = \frac{4}{3} \pi r^3$

$V = 1.124*10^{9}ft^3$

So

$\delta_{iron}=m/V=491lb/ft^3$

$m=V*\delta=5.519*10^{11}lb$

Once you know both masses, you can calculate the force using Newton's universal law of gravitation:

$F_g=G\frac{m_1m_2}{d^2}$

Where G is the gravitational constant:

$G= 1.068846 * 10^{-9} ft^3 lb^{-1} s^{-2}$

$F_g=461lb_f$

4 0

2 years ago

Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts

nignag [31]

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7 0

2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago

the mass of piece of granite is 15.6g when it is suspended can a mass of 5.5g of water is displaced find the dencity of granite

Blababa [14]

Answer:

2836.36 kg/m³

Explanation:

Applying,

Densty of granite(D')/Density of water(D) = weight of granite(W')/weight of water displaced(W)

D'/D = W'/W................... Equation 1

Make D' the subject of the equation

D' = W'D/W............... Equation 2.

From the question,

Given: W' = mg = 9.8(15.6/1000) = 0.15288 N, W = 9.8(5.5/1000) = 0.0539 N, Constant: D = 1000 kg/m³

Substitute these values into equation 2

D' = 1000(0.15288)/0.0539

D' = 2836.36 kg/m³

3 0

1 year ago