<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
Answer:
- gravitational 2.force of gravity
Explanation:
1.it ocours when an object is thrown into the sky. 2.iy ocurs when an object is falling or being pulled from the sky
<span>its kinetic energy is 7350kJ
</span>
Kinetic energy is given as =
Now, m = 12 gms = 0.012 kg
And, velocity = 35 kilometers/second = 35000 m/sec
Kinetic energy is given as =
![\frac{1}{2} 0.012 kg * 35000*35000 m/[tex] s^{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%200.012%20kg%20%2A%2035000%2A35000%20m%2F%5Btex%5D%20s%5E%7B2%7D%20)
= 6
×1225 ×
m/
= 7350 kJ
edson takes 130 minutes of exersise
step by step answer:
10 x 13 = 130
