Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>
Answer:
0.050 m
Explanation:
The strength of the magnetic field produced by a current-carrying wire is given by

where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.
In this problem, we have:
(current in the wire)
(strength of magnetic field)
Solving for r, we find the distance from the wire:

Answer:

Explanation:
The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.
The above formula can be rewritten as follows

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.
If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.
Therefore the electric flux can be written as a function of time

Answer:
The acceleration of the proton is 2.823 x 10¹⁷ m/s²
The acceleration of the electron is 5.175 x 10²⁰ m/s²
Explanation:
Given;
distance between the electron and proton, r = 7 x 10⁻¹⁰ m
mass of proton,
= 1.67 x 10⁻²⁷ kg
mass of electron,
= 9.11 x 10⁻³¹ kg
The attractive force between the two charges is given by Coulomb's law;

where;
k is Coulomb's constant = 9 x 10⁹ Nm²/c²

Acceleration of proton is given by;
F = ma

Acceleration of the electron is given by;
