Answer:
Explanation:
The car is rolling without slipping so Vcm= R×ω = 0.325×49 = 16
Answer:
The arrow rise 44.64 m high assuming air friction is negligible
Explanation:
According to the law of conservation of energy
Kinetic energy = Potential energy at highest
----1
We know that F=kx
So,
Substitute the value in 1

Hence the arrow rise 44.64 m high assuming air friction is negligible
Answer:

Explanation:
The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

But we know that:

And the normal force is given by the sum of the forces in the vertical direction:

Finally, we have:

So, the minimum value for the coefficient of friction is 0.27.
Answer:
The acceleration of the crate is
.
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :

f is frictional force, 

So, the acceleration of the crate is
. Hence, this is the required solution.
Answer:
using Snells law
Oi = angle of incidence = 58.0°
ni = index of refraction of air = 1.0003
nr = index of refraction of glass = 1.47
c = speed of light in vacuum = 3 x 10^8 m/s
Or = angle of refraction = ?
ni(sinOi) = nR (sinOr)
ni( sinOi)/ nR = sinOr
arcsin(ni(sin0i))/nR = Or
arcsin( 1.0003(sin58.0)) / 1.47
Or = 35.25°
Explanation: