Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity
velocity after 5 s is 
Therefore acceleration during these 5 s


therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s


(b)total distance traveled before stoppage


s=136.89 m
Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.
Answer:
2,54 cm are equal to 1 inch
Explanation:
Doing the conversion:
![55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]](https://tex.z-dn.net/?f=55%5Bcm%5D%2A%5Cfrac%7B1%5Binch%5D%7D%7B2%2C54%5Bcm%5D%7D%20%3D21%2C65%5Binch%5D)
Answer:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C