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2 years ago

The cylinder valve is open and the gas is collected and atmospheric pressure

Physics

1 answer:

jok3333 [9.3K]2 years ago

4 0

Less gas will be collected because some of the gases will escape from the open cylinder valve.

Cylinders used to store carbon dioxide will have thicker walls than those of butane because of higher pressures.

<h3>What are compressed gases?</h3>

Compressed gases are gases which are compressed under high pressure in gas cylinders.

Cylinder valves are used to reduce the pressure of the compressed gases and in the process, some of the gas molecules escape.

Since the cylinder valve is open and the gas is collected at atmospheric pressure, less gas will be collected because some of the gases will escape.

Since, the carbon dioxide not liquefy under pressure compared to butane, the cylinders used to store carbon dioxide will have thicker walls than those of butane.

Learn more about compressed gases at: brainly.com/question/518065

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The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang

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If the radio waves transmitted by a radio station have a frequency of 76.5 mhz, what is the wavelength of the waves, in meters?

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Given: Velocity of light c = 3.00 x 10⁸ m/s

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2 years ago

What is the current in a 160V circuit if the resistance is 2Ω? V= I= R =

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3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

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Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :