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3 years ago

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A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

e the pressure is raised to 5.20 atm. What is the volume of the balloon in the liquid nitrogen?a. 97 Lb. 6.7 Lc. 3.6 Ld. 0.25 L

1 answer:

mrs_skeptik [129]3 years ago

7 0

Answer:

0.25 L

Explanation:

$P_1$ = Initial pressure = 1 atm

$T_1$ = Initial Temperature = 20 °C

$V_1$ = Initial volume = 4.91 L

$P_2$ = Final pressure = 5.2 atm

$T_2$ = Final Temperature = -196 °C

$V_2$ = Final volume

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L$

The pressure experienced by the balloon is 0.25 L

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Answer:

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$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

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(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

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One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

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Answer:

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(b). The direction of the first ball after the collision is 44.16° due south of east.

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Given that,

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$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

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$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

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Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

$m_{1}v_{1}=-m_{2}v_{2}$

$v_{1}=-v_{2}$

Put the value into the formula

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Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

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(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

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$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

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