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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Answer:
the less shielding of electrons
Answer:
Force constant will be 1195.85 N/m
Work done will be 1.6859 J
Explanation:
We have given the force, F = 63.5 N
Spring is stretched by 5.31 cm
So x = 0.0531 m
Force is given , F = 63.5 N
We know that force is given by 
So 
k = 1195.85 N/m
Now we have to find the work done
We know that work done is given by
