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)
conduction sdnoajndsojnfojanskfnijoaknfibas
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
Kinetic energy can be passed from one object to another when objects collide,
Answer: True
Hope This Helps! :3
