Answer:
Their total self-inductance assuming they act like a single solenoid is 10.11 μH
Explanation:
Given;
diameter of the heating coil, d = 0.800 cm
combined length of heating coil and hair dryer, = 1.0 m
number of turns, N = 400 turns
Formula for self-inductance is given as;
where
μ₀ is constant = 4π x 10⁻⁷ T.m/A
A is the area of the coil:
A = πd²/4
A = π (0.8 x 10⁻²)²/4
A = 5.027 x 10⁻⁵ m²
Therefore, their total self-inductance assuming they act like a single solenoid is 10.11 μH
Answer:
5.82812 rad/s
Explanation:
L = Length of meter stick = 1 m = 100 cm
= The center of mass of the stick =
= Angular velocity
Moment of inertia of the system is given by
As the energy in the system is conserved
The maximum angular velocity is 5.82812 rad/s
Answer:
the pressure at B is 527psf
Explanation:
Angular velocity, ω = v / r
ω = 20 /1.5
= 13.333 rad/s
Flow equation from point A to B
the pressure at B is 527psf
The horizontal force applied to the block is approximately 1,420.84 N
The known parameters;
The mass of the block, w₁ = 400 kg
The orientation of the surface on which the block rest, w₁ = Horizontal
The mass of the block placed on top of the 400 kg block, w₂ = 100 kg
The length of the string to which the block w₂ is attached, l = 6 m
The coefficient of friction between the surface, μ = 0.25
The state of the system of blocks and applied force = Equilibrium
Strategy;
Calculate the forces acting on the blocks and string
The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N
The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N
Let <em>T</em> represent the tension in the string
The upward force from the string = T × sin(θ)
sin(θ) = √(6² - 5²)/6
Therefore;
The upward force from the string = T×√(6² - 5²)/6
The frictional force = (W₂ - The upward force from the string) × μ
The frictional force, = (981 - T×√(6² - 5²)/6) × 0.25
The tension in the string, T = × cos(θ)
∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
The frictional force on the block W₂, ≈ 219.92 N
Therefore;
The force acting the block w₁, due to w₂ = 219.92/0.25 ≈ 879.68
The total normal force acting on the ground, N = W₁ +
The frictional force from the ground, = N×μ +
= P
Where;
P = The horizontal force applied to the block
P = (W₁ + ) × μ +
Therefore;
P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84
The horizontal force applied to the block, P ≈ 1,420.84 N
Learn more about friction force here;