Answer:
7.1 J
Explanation:
From the question,
Work done by the mover = work done in pushing the crate + work done against friction
W = W'+Wf................. Equation 1
W = mgd+mgμd............ Equation 2
W = mgd(1+μ)................ Equation 3
Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of friction.
Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5
constant: g = 9.8 m/s²
Substitute these values into equation 3
W = 46×9.8×0.0105(1+0.5)
W = 7.1 J
Acceleration is the rate of change of velocity, and velocity is the change in displacement over the change in time so the answer would be A.
The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved.
EXPLANATION :
Before coming into any conclusion, first we have to understand the law of conservation of momentum.
As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.
Hence, during any type of collision, the total momentum is always conserved.
Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.
Answer: The smallest effort = 300N
Explanation:
Using one of the condition for the attainment of equilibrium:
Clockwise moment = anticlockwise moments
900 × 1 = 3 × M
Where M = the weight of the strong man
3M = 900
M = 900/3 = 300N
Therefore, 300N is the smallest effort that the strongman can use to lift the goat
Answer:
The horizontal distance is 0.64 m.
Explanation:
Initial velocity, u =2.5m/s
The maximum horizontal distance is
