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ELEN [110]
3 years ago
12

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 1150 m>min, or 500 ft>min2. What fraction of the engine power is being used to make the airplane climb
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

0.2289

Explanation:

Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then

P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

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An object is allowed to fall freely near the surface of an unknown planet. The object falls 72 meters from rest in 4.0 seconds.
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4 0
3 years ago
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

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\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

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\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

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