Answer:
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
Here d is the distance at which the earthquake take place and 
is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
Answer:
V = V0 + a t
V = 75 - 9.8 * 3 = 45.6 m/s
Current = charge/time = (2 c)/(0.00024 sec)= 8,333 Amps !
