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
Answer:
D
A machine can help decrease the input force and increase the output force.
Answer:
False
Explanation:
The magnitude of any vector is given by,
The magnitude of anything is never negative. It can be even seen from the formula that the components are squared. A squared value can never be negative. Even if the component is negative the square will be always positive.
So, magnitude of the vector is <u>not</u> negative.
Answer:
(a) 2.34 s
(b) 6.71 m
(c) 38.35 m
(d) 20 m/s
Explanation:
u = 20 m/s, theta = 35 degree
(a) The formula for the time of flight is given by
T = 2.34 second
(b) The formula for the maximum height is given by
H = 6.71 m
(c) The formula for the range is given by
R = 38.35 m
(d) It hits with the same speed at the initial speed.
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