91影视

The given data can be listed below as,

• The surface pressure on Venus and on cylindrical benzene tank is 92 atm ,
• Acceleration due to gravity is 0.884 g.
• The diameter of tank is 1.72 m .
• The height of the tank is 11.50 m .

Pascal's law helps to identify the amount of pressure applied to lift or to move an object when a force perpendicular to the surface of the confined liquid is acting on it.

This pressure is equally and evenly distributed in all of the available directions of the liquid.

For the given tank of benzene, density and pressure, conversion are given by,

p of benzene = 90 kg/m3, 1atm = $1.01脳{10}^5\mathrm{Pa}$

By applying Pascal鈥檚 law, pressure can be expressed as,

$p=p_0+pgh$

Here, p is the net pressure, $p_0$ is the atmospheric pressure, p is the density of benzene, g is the acceleration due to gravity , and h is the height of tank

Substitute all the values in the above equation.

Force exerted by the atmosphere of Venus on the outside surface of the bottom of the tank is given by,

$F_{鈯�}=p_{0}A$

Here, $p_0$is the atmospheric pressure of the tank and is the area at the bottom of the tank.

Substitute values in the above equation.

$\begin{array}{r}F=\left(9.32脳{10}^6\mathrm{Pa}\right)蟺\left(\frac{1.72{\mathrm{m}}^2}{4}\right)\\ =2.17脳{10}^7\mathrm{N}\end{array}$

Thus, the force of Venus on the outer surface of the bottom of the water tank is $2.17脳{10}^{7}N$.

The area on vertical walls of the tank is expressed as,

$A=蟺DL$

Here, D is the diameter of the tank and L is the height of the tank.

Substitute all the values above,

$\begin{array}{r}A=\left(1.72\mathrm{m}\right)蟺\left(11.50\mathrm{m}\right)\\ =62.14{\mathrm{m}}^2\end{array}$

According to Pascal鈥檚 law, Atmospheric force exerted on vertical walls of the cylindrical tank is expressed as,

$F_{鈯�}=p_{0}A$

Here, $p_0$is the atmospheric pressure of the tank, and is the area of vertical walls of the tank.

Substitute values in the above equation.

$\begin{array}{r}F=\left(9.32脳{10}^6\mathrm{Pa}\right)\left(62.14{\mathrm{m}}^2\right)\left(\frac{1\mathrm{N}/{\mathrm{m}}^2}{1\mathrm{Pa}}\right)\\ =5.79脳{10}^8\mathrm{N}\end{array}$

Thus, the force on the vertical walls of the tank is $5.79脳{10}^8\mathrm{N}$.