A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
Explanation:
'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.
'Where would you locate the rope to apply the force?' - in point D.
PS. zoom out the attached picture.
Answer:
Follows are the explanation to this question:
Explanation:
In this solution, it is defined that there are two principal motions for the moon, which are its revolution as well as rotation. In such a movement called revolution, its Moon is relocating around the Earth, in which the approximate movement of the moon from around earth has an average movement of about 13.2° per day, or 92 degrees every week, that's once in 27.3 days.
