For the answer to the question above, <span>at the end, the KE = 1/2mv^2 = (1/2)(2000)(2^2) = 4000 J This must equal the net work acting on the car. W=Fd The net force is 1140-950= 190N. so, d=W/F = 4000/190 = 21.05 m
I hope my answer helped you.</span>
Answer:
L = 4.711 *10^{-6} kg m2/s
Explanation:
=4.5*10^-5
angular velocity
= 0.1047 rad/s
the angular momentum,
Answer:
<em>b. At twice the distance, the electric potential is V/2</em>
Explanation:
Electric potential
Is the amount of work needed to move a unit charge from a reference point (usually a point at infinity) without producing acceleration.
The electric potential due to a point charge q at a distance r is given by
Where K is the Coulomb's constant. If we know the electric potential at a certain distance is V, if the distance is changed to 2r, then the new potential is
It means that the electric potential is half the previous value. Correct option: b.
Explanation:
Using kinematics,
s = ut + 0.5at².
We have s = 35.0m, t = 0.50s, a = 2.80m/s².
Substitute in the variables, we have u = 69.3m/s.
This u is only the initial velocity at t1. To find the velocity at t0, we apply a different kinematics equation with different values:
We have t = 5.00s, a = 2.80m/s², v = 69.3m/s.
Using kinematics,
v = u + at.
Substitute in the variables, we have u = 55.3m/s.
Hence the car's initial velocity is 55.3m/s.