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
The area of the bar over r=4 is 0.468
Answer:
Symptoms of excessive stress include all of the following EXCEPT: increased energy level.
Answer:
Force A=-−2,697.75 N
Force B=13, 488.75 N
Explanation:
Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.
25 mg-20Fb=0
25*1100g=20Fb
Fb=25*1100g/20=1375g
Taking g as 9.81 then Fb=1375*9.81=13,488.75 N
The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g
-275*9.81=−2,697.75. Therefore, force A pulls downwards
Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle
The one that accurately describes the products of a reaction is : B. new substances that are present at the end of a reaction
For example the process of photosynthesis transform CO2 and other nutrients into O2 and H2O
hope this helps
