Answer:
a) q = 7.5 10⁻² C
, b) 190 J
, c) I₀ = 50 A
, d) 1.5 mC
Explanation:
The expression for capacitance is
C = q / DV
q = C DV
let's reduce the magnitudes to the SI system
ΔV = 5 kV = 5000 V
C = 15 μF = 15 10⁻⁶ F
t = 90 μs = 90 10⁻⁶ s
q = 15 10⁻⁶ 5000
q = 7.5 10⁻² C
b) the energy in a capacitor is
U = ½ C ΔV²
U = ½ 15 10⁻⁶ 5000²
U = 1,875 10² J
answer 190 J
c) At the moment the discharge begins, all the current is available and it decreases with time,
whereby
V = I R
in the first instant I = Io
I₀ = V / R
I₀ = 5000/100
I₀ = 50 A
but this is for a very short time
answer 50 A
d) The definition of current is
i = dq / dt
in this case they give us the total current and the total time, so we can find the total charge
i = q / t
q = i t
q = 17 90 10⁻⁶
q = 1.53 10⁻³ C
answer is 1.5 mC
Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so

Expression for the heat conduction process is

Expression for the heat convection process is

Substitute the expressions of conduction and convection in equation above


Substitute the values in above equation

Now heat flux through the wall can be calculated as

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Given:
10^10 electrons per second
To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:
From literature,
1 Coulomb is equivalent to 6.242×10^18 electrons<span>.
So,
= 10^10 electrons * (1 coulomb/</span><span>6.242×10^18</span> electrons) / second
<span>= 1.602 x 10^-9 coulumbs
This value is too small to be used in an actual setting.
</span><span>
</span>
Because the drop is moving at constant velocity the net force acting on the raindrop is 0N. Therefore the force of gravity, say XN, pulling the drop downward is canceled out by the force of the air pressing onto the raindrop. So this force will be equal and opposite to the force of gravity, -XN.
When you're talking about gravity, it's easy to identify the equal
opposite forces.
Gravity ALWAYS produces an equal pair of opposite forces.
They both act between the centers of the two objects, one in
each direction.
Consider the equal pair of opposite gravitational forces between
you and the Earth. One force acts on you, and draws you toward
the center of the Earth. We call that force "your weight".
The other one acts on the Earth, and draws it toward the center
of you. Hardly anybody ever talks about that one, but the two
forces are equal ... your weight on Earth is equal to the Earth's
weight on you !