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Nadusha1986 [10]
3 years ago
15

A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.

Physics
1 answer:
babunello [35]3 years ago
8 0
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.


Read more on Brainly.com - brainly.com/question/18743384#readmore
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solution

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Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
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a)

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The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

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The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

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The relationship between torque and angular acceleration \alpha is

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