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Nostrana [21]
4 years ago
7

A quantity of a gas is an absolute pressure of 400 K PA in the absolute temperature of 110° Kelvin when the temperature of the g

as is raised to 235° Kelvin what's the new pressure of the gas?
Physics
1 answer:
aliina [53]4 years ago
4 0

Given:

P1 = 400 kPa
T1 = 110 K

T2 = 235K

Required:

P2

Solution:

Apply Gay-Lussac’s law where P/T = constant

P1/T1 = P2/T2

P2 = T2P1/T1

P2 = (235K)(400kPa) / (110K)

P2 = 855 kPa

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Answer:

F_{3h}=39065.298\times 10^9\ N attractive toward +x axis is the net horizontal force

F_v=80062.47\times 10^9 attractive toward +y axis is the net vertical force

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Given:

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<u>Now the distance between the charge at at origin and the second charge:</u>

d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}

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d_2=0.0435\ m

<u>Now the distance between the charge at at origin and the third charge:</u>

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F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}

F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}

F_2=32175.98\times 10^9\ N attractive towards +y

<u>Now the force due to third charge:</u>

F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}

F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}

F_3=61748.38\times 10^9\ N attractive

<u>Now the its horizontal component:</u>

F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9

F_{3h}=39065.298\times 10^9\ N attractive toward +x axis

<u>Now the its vertical component:</u>

F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9

F_{3v}=47886.49\times 10^9\ N upwards attractive

Now the net vertical force:

F_v=F_{3v}+F_2

F_v=47886.49\times 10^9+32175.98\times 10^9

F_v=80062.47\times 10^9

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