Answer:
The value is [tex]F_{net} = 4444 lb[/tex]
The force will be outward
Explanation:
From the question we are told that
The diameter of the disk is [tex]d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m[/tex]
The external pressure on Mars is [tex]P = 650 \ N/m^2[/tex]
From the question we are told that
Internal pressure = External pressure
Generally the external Force on earth is
[tex]F_E = P_{atm} * A[/tex]
Here [tex]P_{atm}[/tex] is the atmospheric pressure with value [tex]P_{atm} = 1.013*10^{5}\ Pa[/tex]
So
[tex]F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}[/tex]
=> [tex]F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}[/tex]
=> [tex]F_E = 19893 \ N[/tex]
Generally the external Force on Mars is
[tex]F= P * A[/tex]
[tex]F = 650 * \pi * \frac{d^2}{4}[/tex]
=> [tex]F = 650 *3.142 * \frac{0.5^2}{4}[/tex]
=> [tex]F = 127.6 \ N[/tex]
Net force is mathematically represented as
[tex]F_{net} = F_E -F[/tex]
=> [tex]F_{net} = 19893 -127.6[/tex]
=> [tex]F_{net} = 19765.6 \ N[/tex]converting to pounds
[tex]F_{net} = \frac{19765.6}{4.448}[/tex]
=> [tex]F_{net} = 4444 lb[/tex]
Given that that the value is positive then the force will be outward
