The answer to this would be ocean floor
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:
V₂ = -22 V
Explanation:
Electric potential and field are related
ΔV = - E d
where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates
In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m
V₂- V₁ = - E d
V₂ = - Ed + V₁
V₂ = - 4000 0.004 + (- 6)
V₂ = -16 - 6
V₂ = -22 V
Answer:
B. blocks 2 & 3.
Explanation:
Block 1 has equal & opposite forces acting on it.
Block 2 has 5N on one side, 3N on the other. It will move in the direction the 5N of force is pushing.
Block 3 has no opposing force.
