Question

⭕Physics ⭕ 50 points ⭕

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A 30,000 kg air-plane takes off at a speed of 50m/s and 5 min.later it is at an elevation of 3Km with a speed of 100m/s. what average power is required during this 5 min. if 40% of the power is used in overcoming dissipative forces ?​

Answer 1

Answer:

Explanation:

The initial kinetic energy of the plane is:

KEi = (30,000 kg)×(50 m/s)²/2 = 37.5 MJ

The final kinetic energy of the plane is:

KEf = (30,000 kg)×(100 m/s)²/2 = 150 MJ

The change in kinetic energy is:

ΔKE = (150 MJ) - (37.5 MJ) = 112.5 MJ

The initial potential energy of the plane is:

PEi = 0 J

The final potential energy of the plane is:

PEf = (30,000 kg)×(9.8 m/s²)×(3000 m) = 882 MJ

The change in potential energy is:

ΔPE = (882 MJ) - (0 J) = 882 MJ

The total change in energy is:

ΔE = (112.5 MJ) + (882 MJ) = 994.5 MJ

The power is:

p = (994.5 MJ) / (300 s) = 3.315 MW

But that is only 60% of the total power, so

P = (3.315 MW)×100/60 = 5.525 MW < - - - - -

Answer 2

Answer:

Hey mate

Here's ur answer

Explanation:

The initial kinetic energy of the plane is:

KEi = (30,000 kg)×(50 m/s)²/2 = 37.5 MJ

The final kinetic energy of the plane is:

KEf = (30,000 kg)×(100 m/s)²/2 = 150 MJ

The change in kinetic energy is:

ΔKE = (150 MJ) - (37.5 MJ) = 112.5 MJ

The initial potential energy of the plane is:

PEi = 0 J

The final potential energy of the plane is:

PEf = (30,000 kg)×(9.8 m/s²)×(3000 m) = 882 MJ

The change in potential energy is:

ΔPE = (882 MJ) - (0 J) = 882 MJ

The total change in energy is:

ΔE = (112.5 MJ) + (882 MJ) = 994.5 MJ

The power is:

p = (994.5 MJ) / (300 s) = 3.315 MW

But that is only 60% of the total power, so

P = (3.315 MW)×100/60 = 5.525 MW < - - - - - - - - - - - - answer

Hope it helps ❤️