Answer:
Acceleration, 
Explanation:
Given that,
Height from a ball falls the ground, h = 17.3 m
It is in contact with the ground for 24.0 ms before stopping.
We need to find the average acceleration the ball during the time it is in contact with the ground.
Firstly, find the velocity when it reached the ground. So,
u = initial velocity=0 m/s
a = acceleration=g
It is in negative direction, u = -18.41 m/s
Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.
So, the average acceleration of the ball during the time it is in contact is .
Answer:
26b) 66.7%
27) 500 N
Explanation:
26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.
26.b) Mechanical advantage is the ratio between the forces:
MA = load force / effort force
Efficiency is the ratio between the work:
e = work done on load / work done by effort
Work is force times distance.
e = (F load × d load) / (F effort × d effort)
Rearranging:
e = (F load / F effort) × (d load / d effort)
e = MA × (d load / d effort)
In a two pulley system, the load moves half the distance of the effort. So the efficiency is:
e = (4/3) × (1/2)
e = 2/3
e = 66.7%
27) In a three pulley system, the load moves a third of the distance of the effort.
e = (F load / F effort) × (d load / d effort)
0.40 = (600 N / F) × (1/3)
F = 500 N
Light energy is not kinetic energy
Answer:
a) t = 0.528 s
b) D = 1.62 m
Explanation:
given,
speed of the baseball = 3.75 m/s
angle made with the horizontal = 35°
height of the roof edge = 2.5 m
using equation of motion
4.9 t² + 2.15 t - 2.5 = 0
on solving the above equation
t = 0.528 s
b) D = v cos θ × t
D = 3.75 × cos 35° ×0.528
D = 1.62 m
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