Answer:
A baseball (m= 149g) approaches a bat horizontally at a speed of 40.2 m/s (90 mi/h) and is hit straight back at a speed of 45.6m/s (102mi/h). If the ball is in contact with the bat for a time of 1.10ms, what is the average force exerted on the ball by the bat ? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.
Explanation:
Use the impulse equation (a form of Newton's 2nd Law): FΔt = Δ(mv) where Δ means "change in"
The change in momentum is mBB(vf - vi) = (.150 kg)(-46.9 m/s - 40.5 m/s)
Divide this by the time interval and you get F exerted by the bat in Newtons.
Take care.
We have that the speed of a body covering a distance of 320 km in 4h is mathematically given as
V=22.22m/s is
<h3 /><h3>
Speed</h3>
From the question we are told
calculate the speed of a body covering a distance of 320 km in 4h
Generally the equation for the Speed is mathematically given as
V=22.22m/s
Hence
The speed of a body covering a distance of 320 km in 4h is
V=22.22m/s
For more information on Speed visit
brainly.com/question/7359669
What are you asking? And more info needs to be provided.
From the information given, The mass of the bowling ball is 8 Kilograms and the momentum with which it is moving is 16 kg m/s.
We use the formula p = m × v
Where p is the momentum, m is the mass and v is the velocity.
We need velocity so we rewrite the equation thus:
P = mv, therefore p/m = v or v = p/m
In our case p = 16 and m = 8
v = p/m
v = 16/8
v = 2
Therefore the bowling ball is travelling at 2m/s
