Answer:
2156J
Explanation:
Given parameters:
Height of lift = 10m
Mass = 22kg
Unknown:
Work done by the machine = ?
Solution:
Work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Here;
Work done = mass x acceleration due to gravity x height
Work done = 22 x 9.8 x 10 = 2156J
Answer:
![1600 \frac{m}{s^2}](https://tex.z-dn.net/?f=1600%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Explanation:
Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:
![a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bchange-in-velocity%7D%7Btime%7D%20%3D%20%5Cfrac%7BVf-Vi%7D%7Bt%7D)
Where Vf is the final velocity of the object, (in our case 80 m/s)
Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)
and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.
Therefore we have:
![a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B80%20m%2Fs%20-%200%20m%2Fs%7D%7B0.05%20sec%7D%20%3D%201600%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Notice that the units of acceleration in the SI system are
(meters divided square seconds)
Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:
A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
ΔP=20 kg.m/s
Explanation:
Given data
Mass m=0.2 kg
Initial speed Vi=-44.5m/s
Final speed Vf=55.5 m/s
Required
Change in momentum ΔP
Solution
First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:
![v_{i}=-44.5m/s\\v_{f}=55.5m/s](https://tex.z-dn.net/?f=v_%7Bi%7D%3D-44.5m%2Fs%5C%5Cv_%7Bf%7D%3D55.5m%2Fs)
Now we need to find the initial momentum
So
![P_{1}=m*v_{i}](https://tex.z-dn.net/?f=P_%7B1%7D%3Dm%2Av_%7Bi%7D)
Substitute the given values
![P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s](https://tex.z-dn.net/?f=P_%7B1%7D%3D%280.2kg%29%28-44.5m%2Fs%29%5C%5CP_%7B1%7D%3D-8.9kg.m%2Fs)
Now for final momentum
![P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s](https://tex.z-dn.net/?f=P_%7B2%7D%3Dmv_%7Bf%7D%5C%5CP_%7B2%7D%3D%280.2kg%29%2855.5m%2Fs%29%5C%5CP_%7B2%7D%3D11.1kg.m%2Fs)
So the change in momentum is given as:
ΔP=P₂-P₁
![=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s](https://tex.z-dn.net/?f=%3D%5B%2811.1kg.m%2Fs%29-%28-8.9kg.m%2Fs%29%5D%5C%5C%3D20kg.m%2Fs)
ΔP=20 kg.m/s