Answer:
It would take
time for the capacitor to discharge from
to
.
It would take
time for the capacitor to discharge from
to
.
Note that
, and that
.
Explanation:
In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is
, and the initial charge of the capacitor be
. Then at time
, the charge stored in the capacitor would be:
.
<h3>a)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
<h3>b)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
There are creases. Or just origami
Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A =
= 6.77 ×
m²
Now Area of cylinder is :
A =
d²
solving for d:
d = 
d = 9.28 cm