Explanation:
Equilibrium position in y direction:
W = Fb (Weight of the block is equal to buoyant force)
m*g = V*p*g
V under water = A*h
hence,
m = A*h*p
Using Newton 2nd Law

Hence, T time period
T = 2*pi*sqrt ( h / g )
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answer: 
Explanation:
Given
mass of laptop m=2 kg
The velocity of car u=8 m/s
The coefficient of static friction is 
The coefficient of kinetic friction is 
As the car is moving, so the coefficient of kinetic friction comes into play
deceleration offered by friction 
Using the equation of motion 
insert the values
