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boyakko [2]
3 years ago
14

Work is required to compress 5.00 mol of air at 20.00C and 1.00 atm to one-tenth of the original volume by an adiabatic process.

How much work is required to produce this same compression?
Physics
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:46.03 KJ

Explanation:

no of moles(n)=5

temperature of air(T)=20^{\circ}

pressure(p)=1 atm

final volume is \frac{V}{10}

We know work done in adaibatic process is given by

W=\frac{P_iV_i-P_fV_f}{\gamma -1}

\gamma for air is 1.4

we know P_iV_i^{\gamma }= P_fV_f^{\gamma }

1\left ( V\right )^{\gamma }=P_f\left (\frac{v}{10}\right )^{\gamma }

10^{\gamma }=P_f

P_f=25.118 atm

W=\frac{1\times 0.1218-25.118\times 0.01218}{1.4 -1}

W=-46.03 KJ

it means work is done on the system

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We have: Gravitational Potential Energy (U) = mgh
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Substitute their values into the expression:
U = 3 × 9.8 × 3
U = 88.2 J

In short, Your Answer would be 88.2 Joules.

Hope this helps!
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4 years ago
You yell into a canyon. You hear the echo in 3 seconds. How long did the sound of your voice travel before bouncing off a cliff?
Lilit [14]

Answer:

The sound travelled 516 meters before bouncing off a cliff.

Explanation:

The sound is an example of mechanical wave, which means that it needs a medium to propagate itself at constant speed. The time needed to hear the echo is equal to twice the height of the canyon divided by the velocity of sound. In addition, the speed of sound through the air at a temperature of 20 ºC is approximately 344 meters per second. Then, the height of the canyon can be derived from the following kinematic formula:

2\cdot h = v\cdot t (1)

Where:

h - Height, measured in meters.

v - Velocity of sound, measured in meters per second.

t - Time, measured in seconds.

If we know that  v = 344\,\frac{m}{s} and t = 3\,s, then the height of the canyon is:

h = \frac{v\cdot t}{2}

h = \frac{\left(344\,\frac{m}{s} \right)\cdot (3\,s)}{2}

h = 516\,m

The sound travelled 516 meters before bouncing off a cliff.

7 0
3 years ago
How much will an object weigh on the moon if it weighs 60 lbs on the earth?
Xelga [282]

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Explanation:

8 0
3 years ago
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In 1949, an automobile manufacturing company introduced a sports car (the "Model A") which could accelerate from 0 to speed v in
Fudgin [204]

Answer: \frac{P_B}{P_A} = 8.5264

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The energy from the engine is converted into kinetic energy, which is calculated as: KE = \frac{1}{2}.m.v^{2}

To compare the power of the two cars, first find the Kinetic Energy each one has:

<u>K.E. for Model A</u>

KE_A = \frac{1}{2}.m.v^{2}

<u>K.E. for model B</u>

KE_B = \frac{1}{2}.m.(2.92v)^{2}

KE_B = \frac{1}{2}.m.8.5264v^{2}

Now, determine Power for each model:

<u>Power for model A</u>

P_{A} = \frac{m.v^{2} }{2.t}

<u>Power for model B</u>

P_B = \frac{m.8.5264.v^{2} }{2.t}

Comparing power of model B to power of model A:

\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }

\frac{P_B}{P_A} = 8.5264

Comparing power for each model, power for model B is 8.5264 better than model A.

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Ainat [17]
U know by if they are in first place
8 0
3 years ago
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