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Valentin [98]
3 years ago
12

At a pressure 42 kPa, the gas in a cylinder has a volume of 11 liters. Assuming temperature remains the same, if the volume of t

he gas is decreased to 9 liters, what is the new pressure?
Physics
1 answer:
nlexa [21]3 years ago
5 0
Boyle's law<span> is a gas </span>law<span>, stating that the pressure and volume of a gas have an inverse relationship , when temperature is held constant. That is PV = constant. Therefore, (PV)initial = (PV)final. 42x11 = 9x P(final). P(final) = 42x11/9 = 51.34kPa. </span>
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What happens to the direction of an object as an unbalanced force acts on<br> it? Give an example
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Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

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From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

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Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

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Hence, This is the required solution.

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