Answer:
μ = 0.109
Explanation:
Draw a free body diagram of the crate. There are four forces:
Weight force mg pulling down.
Normal force N pushing up.
Applied force P pulling at θ above the horizontal.
Friction force Nμ pushing to the left.
Sum of the forces in the y direction:
∑F = ma
N + P sin θ − mg = 0
N = mg − P sin θ
Sum of the forces in the x direction:
∑F = ma
P cos θ − Nμ = ma
P cos θ − ma = Nμ
μ = (P cos θ − ma) / N
μ = (P cos θ − ma) / (mg − P sin θ)
Given:
P = 585 N
θ = 28.0°
m = 125 kg
a = 3.30 m/s²
μ = (585 cos 28.0° − 125 kg × 3.30 m/s²) / (125 kg × 9.8 m/s² − 585 sin 28.0°)
μ = 0.109
In that case, their momentum must be equal.
So, m1v1 = m2v2
20 * 20 = 40 * v2
v2 = 400 / 40
v2 = 10
In short, Your Answer would be: 10 m/s
Hope this helps!
Answer: B
adding force will add accesion
Answer:
θ = (7π / 3) rad
Explanation:
given,
displacement of simple harmonic motion along x-axis
equation is given as
x = 5 sin (π t + π/3 )
general equation of simple harmonic motion
x = A sin θ
θ is the phase angle
θ = π t + π/3
at t = 2 s
Phase of the motion at t =2 s is θ = (7π / 3) rad
