A21.0-kg block is initially at rest on a horizontal surface. A horizontal force of73.0N is required to set the block in motion,
1 answer:
Answer:
a) μs=0.34
b) μk=0.27
Explanation:
Given that
Mass , m = 21 kg
Force required to move the block ,F= 73 N
The force required to keep the block in motion ,F ' = 58 N
Coefficient of the static friction :
The maximum value of the static friction force
fr= μs m g
At rest position
F= fr
73 = μs x 21 x 10 (take g= 10 m/s²)
μs=0.34
Therefore the Coefficient of the static friction will be 0.34.
Coefficient of the kinetic friction ;
Value of the kinetic friction force ,fr= μk m g
F' = Fr
To move the block at constant speed
58 = μs x 21 x 10 (take g= 10 m/s²)
μk=0.27
Therefore the Coefficient of the static friction will be 0.27.
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Answer:
Explanation:
Check attachment for solution.
Generally the movement of the knight is L, i.e (2,1), (1,2),(-1,2),(1,-2) etc.
So using Pythagoras theorem
x^2+y^2=1^2+2^2
x^2+y^2=5
Then, the knight will make one movement when the displacement is √5.
So let take a look at other positions
(1,0),(0,1),(-1,0), (0,-1).
Then, for us to have this kind of movement, the knight has to make 3 movements.
When the displacement is 1, then it will make 3 movement.
Let examine other positions
(2,2) or(-2,-2) or (2,-2) or (-2,2)
When the displacement is √8
Then, the movement of the knight is 4.
Let examine other points
(2,0) or (0,2) or (-2,0) or (0,-2)
When the displacement is 2.
The knight make 2 movement
