Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
Answer:
47.5kJ
Explanation:
Before climbing the cliff
At 2.
Answer:
I hypothesis that the motion involving the balls in the experiment were moving to create data.
Explanation:
I hope this helps!
Answer:
Explanation:
A. Using
E= ma/q
E=m/q(2s/t²)
So
E= 9.1x10^-31/1.6*10^-19( 2*4.5/ 3*10-12)
E=5.7NC
The electric field has to be downward since the force is positive that is upward
B.
The electron acceleration is of the order of 10^11 times greater so for practical purposes we neglect the effect of gravity
Answer:
19.98Joules
Explanation:
Energy possessed by the body is the kinetic energy
Kinetic Energy = 1/2mv²
m is the mass
v is the velocity
For the 4kg moving at 6m/s
kE = 1/2 * 4 * 6^2
KE = 1/2 * 4 * 36
KE = 72Joules
For the 4kg moving at 5.1m/s
kE = 1/2 * 4 * 5.1^2
KE = 1/2 * 4 *26.01
KE = 52.02Joules
Amount of Energy lost = 72 - 52.02
Amount of Energy lost = 19.98Joules
Hence the amount of Energy lost is 19.98Joules
