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3 years ago

5

A steady device in each cycle converts 155 J of work to thermal energy that is released into a reservoir at 340 K. Part A Calcul

ate the change in entropy of the environment that results from this transfer.

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

change in entropy is 3.3034 × $10^{22}$

Explanation:

give data

thermal energy Q = 155 J

temperature T = 340 K

to find out

change in entropy

solution

we know change in entropy formula that is

change in entropy = Q / ( K×T ) ..............1

here K is boltzmann constant that is 1.38 × $10^{-23}$ kg-m²/s²

put these value in equation 1 we get

change in entropy = Q / ( K × T )

change in entropy = 155 / ( 1.38 × $10^{-23}$ × 340 )

change in entropy = 3.3034 × $10^{22}$

so change in entropy is 3.3034 × $10^{22}$

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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3 years ago

What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?

olya-2409 [2.1K]

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth: (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon: (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference. In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup. Weight is just the force of gravity between the feather and the Earth. It's not affected by what's around the feather, or what's happening to it.

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3 years ago

Write a hypothesis why the moon has very little liquid water.

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3 years ago

Electromagnetic waves require a medium to transfer energy<br> True or False

Simora [160]

The statemengt is <em>false</em>. Electromgnetic waves do not require a medium to transfer energy. They can transfer energy through vacuum (empty space with nothing in it). If electrimagnetic waves needed a medium to move through from place to place, then we could never see the sun, and also it would get very cold here.

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3 years ago

A sound wave has a frequency of 247 hz and a wavelength of 1.4 m. what is the speed of the sound wave in air?

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Speed = wavelength * Frequency
s = 247 /s * 1.4 m
s = 345.8 m/s

In short, Your Answer would be 345.8 m/s

Hope this helps!

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3 years ago