Work=f.d
Work=100*50 = 500 
Power = work/time = 500/4 
=125 watt
        
             
        
        
        
Dispersion im pretty sure
        
             
        
        
        
Answer:
a. V=11.84 m/s
b.x=0.052m
Explanation:
a).
Given 
 ,
, ,
,  .
.







b).

No friction on the ball so:



 
        
             
        
        
        
Answer:
2 m/s
Explanation: Given that a flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe) with a velocity of 6 m/s to the west.
Since the jogger is moving in an opposite direction to the direction of the train, and velocity is the distance covered in a specific direction, the jogger will be moving at a velocity relative to the velocity of the train.
Velocity = (8 - 6) m/s
Velocity = 2 m/s
Therefore, the jogger will be moving at the speed of 2 m/s
 
        
             
        
        
        
Answer: A
Explanation:
From the question, the given parameters are given. 
Mass M = 30kg
Radius r = 2 m
Coefficient of static friction μ = 0.8
Coefficient of kinetic friction μ = 0.6
Kinetic friction Fk = μ × mg
Fk = 0.6 × 30 × 9.8
Fk = 176.4 N
The force acting on the merry go round is a centripetal force F. 
F = MV^2/r
This force must be greater than or equal to the kinetic friction Fk. That is, 
F = Fk 
F = 176.4
Substitute F , M and r into the centripetal force formula above 
176.4 = (30×V^2)/2
Cross multiply 
352.8 = 30V^2
V^2 = 352.8/30
V = sqrt (11.76) m/s
V = 5.24 m/s
Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately