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Rudiy27
3 years ago
5

Air in a cylinder is compressed to one-tenth of its original volume without change in temperature. What happens to its pressure?

Imagine now that a valve is opened in order to restore the initial pressure value. What percentage of the molecules have escaped?
Physics
1 answer:
alexandr402 [8]3 years ago
3 0
<h2>One-tenth of Original Volume</h2>

Explanation:

  • Air in a cylinder is compressed to one-tenth of its original volume without  a change in temperature then the pressure will be increased
  • If the valve is opened in order to restore the initial pressure value the molecules will be escaped
  • The percentage of the molecules that have escaped to attain the initial pressure will be equal to the one-tenth of its original volume
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The answer:
the full question is as follow:
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As shown in the figure, 
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
 D = 0 - (A + B + C) , 
let  W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics

Magnitude
 A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
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4 0
2 years ago
A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
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Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

=>0-2.76=v_1-v_2 ------(B)

From equation (A) and (B)

v_1=-0.739\frac{m}{s}

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3 0
3 years ago
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Answer:

a) The velocity of rock at 1 second, v = 9.8 m/s

b) The velocity of rock at 3 second,  v = 29.4 m/s

c) The velocity of rock at 5.5 second,  v = 53.9 m/s

Explanation:

Given data,

The rock is dropped from a bridge.

The initial velocity of the rock, u = 0

a) The velocity of rock at 1 second,

   Using the first equation of motion

                         v = u + gt

                         v = 0 + 9.8 x 1

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b) The velocity of rock at 3 second,

                         v = u + gt

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c) The velocity of rock at 5.5 second,

                         v = u + gt

                         v = 0 + 9.8 x 5.5

                         v = 53.9 m/s

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Answer:

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V = U + at

Where;

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U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the equation, we have;

V = u + at

V = 10 + 2*4

V = 10 + 8

V = 18 m/s

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