Question

Suppose a distant world with surface gravity of 7.44 m/s2 has an atmospheric pressure of 8.04 3 104 Pa at the surface.

Answer 1

Answer:

1010713.18851 Pa

387999.259089 N

$2.48335668\times 10^9\ Pa$

Explanation:

$P_0$ = Atmospheric pressure = $8.043\times 10^4\ Pa$

r = Radius = 2 m

h = Depth = 10 m

$\rho$ = Density of liquid methane = $415\ kg/m^3$

g = Acceleration due to gravity = 7.44 m/s²

A = Area

Force is given by

$F=P_0A\\\Rightarrow F=8.043\times 10^4\times \pi\times 2^2\\\Rightarrow F=1010713.18851\ N$

The force exerted is 1010713.18851 Pa

$F=mg\\\Rightarrow F=\rho Vg\\\Rightarrow F=415\times \pi 2^2\times 10\times 7.44\\\Rightarrow F=387999.259089\ N$

The weight of the column of methane is 387999.259089 N

The pressure at a depth is given by

$P=P_0+\rho gh\\\Rightarrow P=8.043\times 10^4\times +415\times 7.44\times 10\\\Rightarrow P=2.48335668\times 10^9\ Pa$

The pressure at the depth is $2.48335668\times 10^9\ Pa$