A transcript of a presidential speech (APEX Class ;)
Complete Question
A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.
Answer:
The value for static friction is 
The value for static friction is 
Explanation:
From the question we are told that
The mass of the clock is 
The first horizontal force is 
The second horizontal force is 
Generally the static frictional force is equal to the first horizontal force
So

=> 
=> 
Generally the kinetic frictional force is equal to the second horizontal force
So



what? I guess:
- practice different habits. If you fail don't give up.
- don't always trust people, some are not what they seem.
this question doesn't make any sense...
Answer:
Explanation:
Initial speed, v = 10 x 10^3 m/s
Mass of the earth, M = 6 x 10^24 kg
Radius of the earth, R = 6.4 x 10^6 m
Maximum from the surface of earth, h = ?
Let m = Mass of the projectile
Solution:
Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface



=
=



For me: WASH OUR HANDS REGULARLY