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
Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by
So, 
.... (1)
Now,
Substitute the value of v from equation (1), we get
Let the radius of the alpha particle is r2.
For proton
So, 
... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So, 
... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get
q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values
So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
Answer:
They affected ocean levels, causing waves to affect different areas of the coastline at different times. Lake Eola is a sinkhole lake.
Answer:
Explanation:
For an electric force, F the formula:
F = kQq/r^2
Given:
r2 = 1/2 × r1
F1 × r1 = k
F1 × r1 = F2 × r2
F2 = (F1 × r1^2)/(0.5 × r1)^2
= (F1 × r1^2)/0.25r1^2
= 4 × F1.
