a) 6.52 m/s^2
b) 23.47 km/hr^2
a) v = v0 + at --> 9 = 0 + a(1.38) --> a = 6.52
b) 6.52 m/s2 * 60 s2 / 1 min2 * 60 min2 / 1 hr2 = 23.47 km/hr2
The stretch of the OA spring is calculated by the formula of F=-k*(extension) and thats the formula for elasticity. Based on my calculation i came up with an answer of the unstretched length is 2m abd the stiffnes is 310n/m
Answer:
Explanation:
Impulse of a force is measured by force x time or F X t
Impulse also equals change in momentum or
F x t = m v₂ - m v₁
The given case is as follows
in the first case
F x t = mv - o = mv
F = mv / t
in the second case
F₁ x 4 t = mv
F₁ = 1/4 x mv /t
F₁ = F / 4
option a) is correct .
iii )
In the last case
F₂ X t = m v/2 -0
F₂ = 1/2 x mv / t
= 1/2 x F
F₂ = F/2
Option e ) is correct.
The net force with be 10 N east. As both boys are pushing a wagon to the same direction, you just add the numbers together. If the directions were opposite like if it was the boy pushing 6 N east and the other boy pushing 4 N west...then it would come out as 2 N east (you subtract and go to the direction with the biggest number) if you have any confusions, feel free to ask
To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as
Here,
k = Spring constant
m = Mass
Our values are given as,
Rearranging to find the spring constant we have that,
Therefore the spring constant is 1.38N/m
