Equation of velocity is given as
![V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^](https://tex.z-dn.net/?f=V%20%3D%5B5.00m%2Fs%E2%88%92%280.0180m%2Fs%5E3%29t%5E2%5Di%5E%20%2B%20%5B2.00m%2Fs%2B%280.550m%2Fs%5E2%29t%5Dj%5E)
at t = 7.93 s
so the magnitude of the velocity is given as
Part b)
the direction of the velocity is given as
part c)
for acceleration we know that
at t = 7.93 s
magnitude is given as
Part d)
for the direction of the motion
Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
<span>0.166666666667
Hope this helped!
STSN</span>
Answer:
972 J
Explanation:
At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:
GPE = mass×gravity×heigth
GPE = 2.2×9.8×45.08 ≈ 972
The impulse exerted by a force F on an object in a time
is given by
In our problem, we have a force of 120 N: F=120 N, applied to the baseball for a time of
. Therefore the impulse is
