It mainly travels by kinetic energy
Solution :
Given :
M = 0.35 kg
Total mechanical energy = constant
or 
But 
and 
Therefore, potential energy at the top = kinetic energy at the bottom
(h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :
here, 
∴ 
= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
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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Explanation:
If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.
Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.
