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Papessa [141]
2 years ago
13

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

c pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 4.4 × 10-4 m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.26 × 105 Pa. For this problem, assume that the density of air maintains a constant value of rhoa = 1.20 kg / m3 and that the density of seawater maintains a constant value of rhos = 1025 kg / m3.What is the maximum height h in meters above the ground that the container can be lifted before bursting?
Physics
1 answer:
MaRussiya [10]2 years ago
7 0

Answer:

dz=19217687.07\ m

Explanation:

Given:

  • initial gauge pressure in the container, P_0=2.02\times 10^{5}\ Pa
  • atmospheric pressure at sea level, P_a=1.01\times 10^5\ Pa
  • initial volume, V_0=4.4\times 10^{-4}\ m^3
  • maximum pressure difference bearable by the container, dP_{max}=2.26\times 10^{5}\ Pa
  • density of the air, \rho_a=1.2\ kg.m^{-3}
  • density of sea water, \rho_s=1.2\ kg.m^{-3}

<u>The relation between the change in pressure with height is given as:</u>

\frac{dP_{max}}{dz} =\rho_a.g_n

where:

dz = height in the atmosphere

g_n= standard value of gravity

<em>Now putting the respective values:</em>

\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8

dz=19217.687\ km

dz=19217687.07\ m

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

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