Respuesta :
Answer:
[tex]dz=19217687.07\ m[/tex]
Explanation:
Given:
- initial gauge pressure in the container, [tex]P_0=2.02\times 10^{5}\ Pa[/tex]
- atmospheric pressure at sea level, [tex]P_a=1.01\times 10^5\ Pa[/tex]
- initial volume, [tex]V_0=4.4\times 10^{-4}\ m^3[/tex]
- maximum pressure difference bearable by the container, [tex]dP_{max}=2.26\times 10^{5}\ Pa[/tex]
- density of the air, [tex]\rho_a=1.2\ kg.m^{-3}[/tex]
- density of sea water, [tex]\rho_s=1.2\ kg.m^{-3}[/tex]
The relation between the change in pressure with height is given as:
[tex]\frac{dP_{max}}{dz} =\rho_a.g_n[/tex]
where:
dz = height in the atmosphere
[tex]g_n[/tex]= standard value of gravity
Now putting the respective values:
[tex]\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8[/tex]
[tex]dz=19217.687\ km[/tex]
[tex]dz=19217687.07\ m[/tex]
Is the maximum height above the ground that the container can be lifted before bursting. (Since the density of air and the density of sea water are assumed to be constant.)