When its engine of power 75.0 kW is generating full power, a small single-engine airplane with mass 690 kg gains altitude at a rate of 2.10 m/s.
What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)
g=9.80, express answer as a percentage
2006-06-28 19:37:17 · 8 answers · asked by Ronaldinho G 1 in Science & Mathematics ➔ Physics
18.67%
2006-06-28 19:45:52 · answer #1 · answered by happyharrytick 3 · 0⤊ 0⤋
1
2016-06-09 07:52:43 · answer #2 · answered by Orpha 3 · 0⤊ 0⤋
First of all, the work the airplane does can be calculated as follows:
Work done = force x distance
=mass x acceleration x distance
However, we are not given distance but its speed, that is the distance it travels in one second
=690 Kg x 9.80 ms^-2 x 2.10 ms^-1
=14200.2 W, that is 14.2 KW < approx >
The % of power the plane uses to ascend is therefore
= ( 14.2 / 75 ) x 100%
=18.93... % = 18.9 % <3SF>
2006-07-02 08:09:51 · answer #3 · answered by sexy_blue7 2 · 0⤊ 0⤋
Easy. The power consumed by the gain in altitude is
P = 690 * 9.8 * 2.10 watts
or around 14,000 watts.
That's less than 20% of 75,000. DrawABlank is wrong. Robert F is an idiot. Happy Harry Tick could be right.
2006-06-28 19:40:16 · answer #4 · answered by ? 6 · 1⤊ 0⤋
When the aircraft is climbing, it is gaining potential energy of m*g*h.
If h is increasing at x per sec, the energy rate (power) is m*g*x
So 690kg * 9.8 m/s^2 * 2.1 m/s gives you the power used to make the airplane climb. This result is 14,200.2 joules/sec which is watts. This is 18.9% of 75kw
2006-06-28 19:56:33 · answer #5 · answered by gp4rts 7 · 0⤊ 0⤋
3%
2006-06-28 19:41:34 · answer #6 · answered by Robert F 7 · 0⤊ 0⤋
Well now, that is a good question. Not? 69049832348569555%
2006-06-28 19:51:18 · answer #7 · answered by grannywinkie 6 · 0⤊ 0⤋
25.35975445223699852155226688412%
2006-06-28 19:41:51 · answer #8 · answered by JZX 4 · 0⤊ 0⤋