Answer:
a) 
b) 
Explanation:
Given:
Mass of the cabinet, m = 51 kg
a) Applied force = 200 N
Now the force required to move the from the state of rest (F) = coefficient of static friction (
)× Normal reaction(N)
N = mg
where, g = acceleration due to gravity
⇒N = 51 kg × 9.8 m/s²
⇒N = 499.8 N
thus, 200N = × 499.8 N
⇒ = 
b) Applied force = 100 N
since the cabinet is moving, thus the coefficient of kinetic(
) friction will come into action
Now the force required to move the from the state of rest (F) = coefficient of kinetic friction ()× Normal reaction(N)
N = mg
where, g = acceleration due to gravity
⇒N = 51 kg × 9.8 m/s²
⇒N = 499.8 N
thus, 100N = × 499.8 N
⇒ = 
1. 2+0.5+2.5= 3. 2km/hr average
2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration
3. 15m/s divided by 2.5 = 6m/s acceleration
Acceleration = v/ t = - 25/5 = - 5 m/s^2 . Minus because object is deaccelerating. A is the correct answer.
Answer:
They preserve many good_from historyof nepal_for us(from history of nepal)
Given:
Mass of the rail road car, m = 2 kg
velocity of the three cars coupled system, v' = 1.20 m/s
velocity of first car, 
= 3 m/s
Solution:
a) Momentum of a body of mass 'm' and velocity 'v' is given by:
p = mv
Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:
(1)
where,
= velocity of the first car
= velocity of the 2 coupled cars after collision
Now, from eqn (1)
v' = 1.80 m/s
Therefore, the velocity of the combined car system after collision is 1.80 m/s
