The solution for this problem is:
Let u denote speed.
Equating momentum before and after collision:
= 0.060 * 40 = (1.5 + 0.060) u
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
Answer:
α = 3×10^-5 K^-1
Explanation:
let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.
then, the coefficient of linear expansion is given by:
α = ΔL/(ΔT×L)
= (0.3×10^-3)/(100)(100×10^-3)
= 3×10^-5 K^-1
Therefore, the coefficient of linear expansion is 3×10^-5 K^-1
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
I think analog but I could be wrong
b) 4m/s/s
This is because you divide the speed you reach, by the time it takes to get to that speed:
12m/s ÷ 3s = 4m/s/s
The units come from what you divide, meters per second ÷ seconds this can be written as m/s/s or ms-²
