Consider a 145 gram baseball being thrown by a pitcher. The ball approaches the batter with a speed of 44 m/s. The batter swings and sends the ball flying back at a speed of 90 m/s into the outfield. If the ball was in contact with the bat for a time of 3 ms, what is the average force exerted by the bat during the collision?
1 answer:
For the development of this problem it is necessary to apply the concepts related to Force, momentum and time.
By definition we know that momentum can be expressed as
Where
P = Momentum
F = Force
t = Time
At the same time, another way of expressing the momentum is through mass and speed, that is,
P = mv
Where
m = Mass
v = Velocity
Equating both equations
F*t = mv
Our values are given as
Net velocity
Replacing we have then,
Therefore the average force exerted by the bat during the collision is 2223.3N
You might be interested in
Answer:
The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.
Explanation:
ω1= 1.72x10^4 rad/sec
ω2= 5.42x10^4 rad/sec
ωmax= 8.42x10^4 rad/sec
θ= 1.72x10^4 rad
α=7.67 x10^4 rad/sec²
t= ωmax / α
t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²
t=1.097 sec
Answer:
<h2>
3 0 0 0 J </h2>
Option C is the correct option .
Explanation:
Given ,
Force = 6 0 0 N
Distance = 5 meters
Work = ?
Now,
Work = Force distance
Calculate the pro duct
Joule
Hope this helps. . .
Good luck on your assignment . .
Weight = (mass) x (gravity) 120 N = (mass) x (9.8 m/s²) Mass = (120 N) / (9.8 m/s²)Mass = 12.24 kg (B)
The length of its image =
Given,
length of object,
We know, for flat refracting surface,
Image distance = object distance
So,
magnification is =
length of the image,
Here, negative sign means inverted image.
For more information on refraction , visit
brainly.com/question/14760207?referrer=searchResults
Answer:
Explanation:
The work increased the potential energy
W = PE = mgh = 40(9.8)(15) = 5880 J(oules)