Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
Answer:
The inductance of solenoid A is twice that of solenoid B
Explanation:
The inductance of a solenoid L is given by
L = μ₀n²Al where n = turns density, A = cross-sectional area of solenoid and l = length of solenoid.
Given that d₁ = 2d₂ and l₂ = 2l₁ and d₁ and d₂ are diameters of solenoids A and B respectively. Also, l₁ and l₂ are lengths of solenoids A and B respectively.
Since we have a cylindrical solenoid, the cross-section is a circle. So, A = πd²/4.
Let L₁ and L₂ be the inductances of solenoids A and B respectively.
So L₁ = μ₀n²A₁l₁ = μ₀n²πd₁²l₁/4
L₂ = μ₀n²A₂l₂ = μ₀n²πd₂²l₂/4
Since d₁ = 2d₂ and l₂ = 2l₁, sub
L₁/L₂ = μ₀n²πd₁²l₁/4 ÷ μ₀n²πd₂²l₂/4 = d₁²/d₂² × l₁/l₂ = (2d₂)²/d₂² × l₁/2l₁ = 4d₂²/d₂² × l₁/2l₁ = 4 × 1/2 = 2
L₁/L₂ = 2
L₁ = 2L₂
So, the inductance of solenoid A is twice that of solenoid B
The volume increases, but the surface area remains the same
Answer:
Gravitational force is directly proportional to the distance between objects.
