Answer:
7.08 m/s²
Explanation:
Given:
v₀ = 20.0 m/s
v = 105 m/s
t = 12.0 s
Find: a
v = at + v₀
105 m/s = a (12.0 s) + 20.0 m/s
a = 7.08 m/s²
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
a
Explanation:
as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it
from rarefaction to rarefaction for a longitudinal wave
Answer:3.31m/s², to the right
Explanation:
According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.
Momentum = mass × velocity
M1 and M2 be the masses of the first and second skaters respectively
Let u1 and u2 be the velocities of the first and second skaters respectively.
v be their common velocity after collision
M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²
According to the law we have
M1u1 + M2u2 = (M1+M2)v
77(4) + 66(2.5) = (77+66)v
308 + 165 = 143v
V = 473/143
V = 3.31m/s²
Their velocity after collision will become 3.31m/s²
They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.
