K=0.5 mu×u
K=2200J no matter the direction
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:
"C" I think....
Explanation:
I am really sorry if I am wrong, but if right, I hope this helps!
In order to calculate the electric field strength, we may use the formula:
E = kQ/d²
Where Q is the charge and d is the distance between the charge and the test charge. Substituting the values into the equation:
E = (9 x 10⁹)(8.7 x 10⁻⁹) / (3.5²)
E = 6.39 Newtons per coulomb
Therefore, the answer is 6.4 Newtons/coulomb