Answer:

Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,

v = 2v

..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,


So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.
I think the answer is periodic motion.
Answer: (A) 3.0=A
Explanation: In order to explain this problem we have to use the OHM law, given by: V=R*I
Besides, we have to consider the resitance equivalent for a parallel connection. This is given by:
1/Re=1/R1+1/R2
If we connect the same resistance, the equivalent resistance is R/2.
Initlally the current is 1.5 A when one resistance is connected to the batttery. When a second resistance with the same value is connected in parallel to the battery, we have V=Re*Ifinal= (R/2)*Ifinal
also we know that V=R*Iinitial so Iinitial=V/R
then Ifinal= 2*V/R=2*Iinitial
Answer:
because potentil energy is redy to go but its bound up
And kinetic energy is in motion
Explanation:
Hi there!
We can use the following (derived) equation to solve for the final velocity given height:
vf = √2gh
We can rearrange to solve for height:
vf² = 2gh
vf²/2g = h
Plug in the given values (g = 9.81 m/s²)
(13)²/2(9.81) = 8.614 m
We can calculate time using the equation:
vf = vi + at, where:
vi = initial velocity (since dropped from rest, = 0 m/s)
a = acceleration (in this instance, due to gravity)
Plug in values:
13 = at
13/a = t
13/9.81 = 1.325 sec