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netineya [11]
3 years ago
15

When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rat

e of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb
Physics
1 answer:
svlad2 [7]3 years ago
8 0

Answer:

343/1500

Explanation:

Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).

From the question,

P' = mg×v................. Equation 1

Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.

Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²

Substitute into equation 1

P' = 700(2.5)(9.8)

P' = 17150 W.

If the full power generated by the engine = 75000 W

The fraction of the engine power used to make the climb = 17150/75000

= 343/1500

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Answer:

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Explanation:

Given that,

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B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

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7 0
3 years ago
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d=412.5 m-x since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance x towards the cliff

t=\frac{2.2 s}{2}=1.1 s Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

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3 years ago
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