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Scorpion4ik [409]
3 years ago
6

Which changes in an electric motor will make the motor stronger? Check all that apply. using a stronger permanent magnet using a

weaker permanent magnet increasing current in the electromagnet decreasing current in the electromagnet increasing the distance between the magnets decreasing the distance between the magnets
Physics
2 answers:
wariber [46]3 years ago
7 0
<h2>Answer:</h2>

The following changes will make the motor more stronger

  • <u>Using strong permanent magnets</u>
  • <u>Increasing the current</u>
  • <u>Decreasing the space between magnets</u>
<h2>Explanation:</h2>

Using string magnets will produce strong flux and the motor will become strong. The internal flux also deepness upon the strength of coil. Similarly by increasing current we can make the motor more powerful because according to power equation P=VI so increasing the current will increase the product of voltage and current resulting more power. In the same way by reducing the distance between magnets will make the motor stronger because the flux and torque will increase.

miv72 [106K]3 years ago
4 0
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
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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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0.273 liters are needed to accomplish this task without boiling.

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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

0.273 liters are needed to accomplish this task without boiling.

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