How much brighter is a sun-like star than the reflected light from a planet orbiting around it?.
1 answer:
You might be interested in
Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
Answer:
a) ω=n
b) V = R*n
c) a = 0
d)
Explanation:
Angular velocity is given by the derivative of θ:
ω = n
Linear Velocity will be:
V = ω*R = n*R
Tangential acceleration will be the derivative of the linear velocity. Since velovity is constant:
a = 0
Radial acceleration is given by: