Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From he question we are told that
The first mass is 
The second mass is 
From the question we can see that at equilibrium the moment about the point where the string holding the bar (where
are hanged ) is attached is zero
Therefore we can say that

Making x the subject of the formula



Looking at the diagram we can see that the tension T on the string holding the bar where
are hanged is as a result of the masses (
)
Also at equilibrium the moment about the point where the string holding the bar (where (
) and
are hanged ) is attached is zero
So basically


Making
subject


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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