We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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Answer:
3125 N
Explanation:
diameter /2 =radius
so r1 =14cm , r2 =35cm
f1/A1 =f2/A2.
f2 = f1 × A2 / A1
=500×1225 pi cm² / 96 pi cm²
f2 =3125N
The average rate at which the cable does work is 294,000 J/s.
The given parameters:
- <em>mass, m = 3000 kg</em>
- <em>height, h = 200 m</em>
- <em>time of motion, t = 20 s</em>
The average rate at which the cable does work is calculated as follows;
Thus, the average rate at which the cable does work is 294,000 J/s.
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Answer:
13.1
Explanation:
that's what I'm gonna go with, but u can research more
Answer:
The movement of an object depends on the reference frame, so it is important to predicate it.
Explanation:
