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Ответ:
The value is![A = 39315 \ m^2](/tpl/images/1005/4222/221a6.png)
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is![v = 1 \ m/s](/tpl/images/1005/4222/19b8e.png)
The mass of the rover and the parachute is![m = 2270 \ kg](/tpl/images/1005/4222/45034.png)
The drag coefficient is![C__{D}} = 0.5](/tpl/images/1005/4222/e2048.png)
The atmospheric density of Earth is![\rho = 1.2 \ kg/m^3](/tpl/images/1005/4222/37d32.png)
The acceleration due to gravity in Mars is![g_m = 3.689 \ m/s^2](/tpl/images/1005/4222/34b26.png)
Generally the Mars atmosphere density is mathematically represented as
=>![\rho_m = 0.71 * 1.2](/tpl/images/1005/4222/08feb.png)
=>![\rho_m = 0.852 \ kg/m^3](/tpl/images/1005/4222/3bb37.png)
Generally the drag force on the rover and the parachute is mathematically represented as
=>
=>
Gnerally this drag force is mathematically represented as
Here A is the frontal area
So
=>![A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }](/tpl/images/1005/4222/7e0be.png)
=>![A = 39315 \ m^2](/tpl/images/1005/4222/221a6.png)