The ballerinas takes advantage of the moment of interia by not stretching her hands out etc. and by doing so decreasing the moment of interia and therefore a smaller torque is produced to rotate at a faster rate.
Answer:
The impulse exerted by one cart on the other has a magnitude of 4 N.s.
Explanation:
Given;
mass of the first cart, m₁ = 2 kg
initial speed of the first car, u₁ = 3 m/s
mass of the second cart, m₂ = 4 kg
initial speed of the second cart, u₂ = 0
Let the final speed of both carts = v, since they stick together after collision.
Apply the principle of conservation of momentum to determine v
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 3 + 0 = v(2 + 4)
6 = 6v
v = 1 m/s
Impulse is given by;
I = ft = mΔv = m(
The impulse exerted by the first cart on the second cart is given;
I = 2 (3 -1 )
I = 4 N.s
The impulse exerted by the second cart on the first cart is given;
I = 4(0-1)
I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).
Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.
Answer:
F = N*μ or F =m*g*μ
Explanation:
The friction force is defined as the product of the normal force by the corresponding friction factor.
When a body is in equilibrium over a horizontal plane its normal force value shall be equal to:
![N = m*g\\where:\\m=mass [kg]\\g=gravity [m/s^2]\\N= normal force [N]](https://tex.z-dn.net/?f=N%20%3D%20m%2Ag%5C%5Cwhere%3A%5C%5Cm%3Dmass%20%5Bkg%5D%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5CN%3D%20normal%20force%20%5BN%5D)
if we simplify this formula more for a balanced body on a horizontal plane, we will have.
==> Object A travels for 60 seconds before Object B starts out.
==> Object A moves at 2 m/s.
==> So Object A has a lead of 120 m when Object B starts out.
==> Object B moves at 3 m/s . . . 1 m/s faster than Object A.
==> So Object B catches up on Object A by 1 m every second.
==> Object B closes up Object A's lead of 120 m in <em>120 seconds</em>.
