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kicyunya [14]
3 years ago
14

A capacitor is charged to 8.0×10−4 C , then discharged by connecting a wire between the two plates. 35us after the discharge beg

ins, the capacitor still holds 14 % of its original charge.
What was the average current during the first 35us of the discharge?
Physics
1 answer:
Anettt [7]3 years ago
7 0

Answer:

The average current is 19.567 A

Solution:

As per the question:

Charge, Q = 8.0\times 10^{- 4}\ C

Time, t = 35\times 10^{- 6}\ s

Now,

We know that current is constituted by the rate of transfer of the charge per unit time. Thus we can write:

I = \frac{Q}{t}                 (1)

Now, the charge that was transferred is 86 % of the original value.

Therefore,

We replace Q by 0.86Q in eqn (1):

I = \frac{0.86\times 8.0\times 10^{- 4}}{35\times 10^{- 6}} = 19.657\ A

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3 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
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Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

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A truck is carrying a steel beam of length 13.0 m on a freeway. An accident causes the beam to be dumped off the truck and slide
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Answer:

emf= 9.88 \times 10^{-3} T

Explanation:

Given:

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