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4 years ago

5

Juan makes an adjustment to an electromagnet that causes the electromagnet to lose some of its strength. What did Juan most like

ly do?

2 answers:

slamgirl [31]4 years ago

8 0

Explanation:

The electromagnet is made by wrapping the coils around the soft iron. Then, pass the current through it. In this way, It will act as a temporary magnet. More the number of coils then, there will be more magnetism.

As long as the current passes through it it will retain its magnetism. When the current is stopped flowing in it then it will lose its magnetism.

In the given problem, Juan makes an adjustment to an electromagnet that causes the electromagnet to lose some of its strength. He can do this by reducing the number of loops in the wire.

Assoli18 [71]4 years ago

5 0

Reducing the amount of loops will cause a loss of strength, as the loops make the magnet stronger.

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3 years ago

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A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic

vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 $m^{2}$ </h3>

Explanation:

Given :

No. of turns in the first coil $N_{1} = 41$

No. of turns in the second coil $N_{2} = 123$

Area of first coil $A_{1} = 0.074 m^{2}$

According to the law of electromagnetic induction,

Induced emf = $-N \frac{d \phi}{dt}$

Where $\phi =$ magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

$-\frac{N_{1} d\phi _{1} }{dt_{1} } = -\frac{N_{2} d\phi _{2} }{dt_{2} }$

$\frac{N_{1} A_{1} dB_{1} }{dt_{1} } = \frac{N_{2} A_{2} dB_{2} }{dt_{2} }$

$A_{2} = \frac{N_{1} A_{1} }{N _{2} }$

$A_{2} = \frac{41 \times 0.074 }{123 }$

$A_{2} = 0.0246 = 0.025 m^{2}$

Therefore, the area of second coil is ≅ 0.025 $m^{2}$

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4 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

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