91影视

The given information is listed below,

• Standard atmospheric pressure is applied on one side of the door
• The difference between the pressure on another side of the door is 1% .

At rest, a surface in a fluid obeys Pascal's law. For this surface and fluid to remain at rest, the fluid must exert forces of equal magnitude in opposite directions on both sides of the surface.

The pressure varies with weather changes and with elevation.

According to Pascal鈥檚 law, the force can be given by,

$F_{鈯�}=p{A}_1$

Here, $p$is the atmospheric pressure on the object at a point whose value is $\left(1.01脳{10}^5\mathrm{Pa}\right)$and $A_1$is the area of the surface.

Let the door has an area (A) of 1 square meter. The new area can be given as,

$$\begin{gathered} A_1=1\%脳A \\ =1\%\left(\frac{1}{100\%}\right)A \\ =0.01脳A \end{gathered}$$

Substitute all the values in the above equation. The force on the door can be calculated as,

$$\begin{gathered} F_{鈯�}=\left(1.01脳{10}^5\mathrm{Pa}\right)\left(0.01\right)\left(1{\mathrm{m}}^2\right)\left(\frac{1\mathrm{N}/{\mathrm{m}}^2}{1\mathrm{Pa}}\right) \\ {F}_{鈯�}=1010\mathrm{N} \end{gathered}$$

This force is approximately equal to the weight of 1 large man (about 102 kg).

Thus, one could push and open the door but with some difficulty.