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alekssr [168]
3 years ago
10

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

external pressure. The hatch is a disk 50.0 cm in diameter. When the lander goes to Mars, where the external pressure is 650 N/m2, what will be the net force (in newtons and pounds) on the hatch, assuming that the internal pressure is the same in both cases? Will it be an inward or outward force?
Physics
1 answer:
Kaylis [27]3 years ago
7 0

Answer:

The value is  F_{net} =  4444 lb

The force will be  outward

Explanation:

From the question we are told that

   The diameter of the disk is  d = 50.0 \ cm  = \frac{50}{100} = 0.5 \ m

   The external pressure on Mars  is P = 650 \ N/m^2

From the question we are told that  

   Internal pressure  =  External pressure

Generally the external Force on earth is

      F_E = P_{atm} * A

Here P_{atm} is the atmospheric pressure with value  P_{atm} = 1.013*10^{5}\ Pa

So

      F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}

=>   F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}

=>   F_E = 19893 \  N

Generally the external Force on Mars is  

       F= P * A

      F = 650 * \pi * \frac{d^2}{4}

=>   F = 650 *3.142 * \frac{0.5^2}{4}

=>   F = 127.6 \  N    

Net force is mathematically represented as

      F_{net} = F_E -F

=>    F_{net} =  19893  -127.6

=>    F_{net} =  19765.6 \ Nconverting to  pounds

    F_{net} = \frac{19765.6}{4.448}

=> F_{net} =  4444 lb

Given that that the value is positive then the force will be  outward

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Part A

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The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

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Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

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ΔQ = W + ΔU

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The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

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The work done by the piston, W = 1,000 J

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V₂ = The final volume

V₁ = The initial volume

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The change in volume ΔV = 1 m³

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