Answer:
Explanation:
It is a question relating to charging of capacitor . For charging of capacitor , the formula is as follows.
Q = CV ( 1 - )
λ = 1/CR , C is capacitance and R is resistance.
= 1/(500 x 10⁻⁶ x 20 x 10³ )
= .1
λ t = .1 x 20
λ t = 2
CV = 500 X 10⁻⁶ X 5
= 2500 X 10⁻⁶ C
Q = 2500 x 10⁻⁶ ( 1 - )
= 2500 x 10⁻⁶ x .86566
= 2161.66 μ C .
voltage = Charge / capacitor
2161.66 μ C / 500μ F
= 4.32 V
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;