The force acting on the ball are unbalanced. Reactionary momentum force (that originated as a result of the swing of the bat) is the most powerful.
Yes friction is acting on the ball. In course of journey it would slow the ball down and make it trace a parabolic path rather than straight path as intended by hitter.
Explanation:
As the hitter hits the ball, momentum of the bat due to swing (mass of the bat*velocity provided by the batsman swinging action of bat) gets transferred on the ball on its impact with the bat.
Since ball’s mass is quite small as compared to the bat, the velocity of the ball increases by the same factor by which the ball’s mass is lower than the bat’s mass. This velocity causes forward motion of the ball (of course in the direction of bat’s motion, here the batsman intends to send the ball straight away hence the ball would move straight).
Various forces on ball is-
- Reactionary momentum force -bat’s force (most powerful force)
- The frictional force of the air (opposing the motion of the ball through the air)
- Gravity force (pulling the ball down to the Earth)
As a combined effect of these force when all the force remains unbalanced, the ball moves away in the straight path under the impact of bats momentum which was most powerful of all.
Frictional force and Gravity force continue acting on the ball. While frictional forces decrease the ball velocity through the air, gravity force pulls it down thus deflecting its direction. Under the combined impact of declining bats momentum, friction force and gravity force, the ball traces a parabolic path (in accordance with the first law of motion from Newton)
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Answer:
(a) 0.177 m
(b) 16.491 s
(c) 25 cycles
Explanation:
(a)
Distance between the maximum and the minimum of the wave = 2A ............ Equation 1
Where A = amplitude of the wave.
Given: A = 0.0885 m,
Distance between the maximum and the minimum of the wave = (2×0.0885) m
Distance between the maximum and the minimum of the wave = 0.177 m.
(b)
T = 1/f ...................... Equation 2.
Where T = period, f = frequency.
Given: f = 4.31 Hz
T = 1/4.31
T = 0.23 s.
If 1 cycle pass through the stationary observer for 0.23 s.
Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.
= 16.491 s.
(c)
If 1.21 m contains 1 cycle,
Then, 30.7 m will contain (30.7×1)/1.21
= 25.37 cycles
Approximately 25 cycles.
The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N, 
= 25 N, a = -0.9 
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.
= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
Answer:
h2 = 0.092m
Explanation:
From a balance of energy from point A to point B, we get speed before the collision:
Solving for Vb:
Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:
Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:
Solving for h2:
h2 = 0.092m
