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3 years ago

What is an advantage of using an electromagnet rather than a regular magnet

Physics

2 answers:

Lena [83]3 years ago

5 0

Electromagnet would have a stronger pull than a regular magnet
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SVEN [57.7K]3 years ago

4 0

Answer:

Electromagnets are temporary magnets made by supplying electric current to a soft iron wounded with coils of wire. Once the source of electricity is disconnected, the magnetism dies off.

However electromagnets usually have stronger electric fields than regular magnets, therefore they are often used in lifting large piece of equipment from one point to another, provided that the equipment are magnetic in nature.

Another major advantage of using an electromagnet rather than a regular magnet is that the magnetism of the electromagnet can be terminated when need be by simply disconnecting the source of electricity.

For example in lifting a heavy metal from one point to another if it gets to the point of dropping off the metal there will be a need to terminate the magnetism so that the metal can fall off; a regular magnet can not be used for this purpose.

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Answer:

Element or a compound.

Explanation:

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Suppose that there are two very large reservoirs of water, one at a temperature of 91.0 °C and one at a temperature of 17.0 °C.

stepan [7]

Answer:32.83

Explanation:

Given

$T_H=91^{\circ}\approx 364 K$

$T_L=17^{\circ}\approx 290 K$

Q=46830 J

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3 years ago

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What is the effective resistance between the points A and D? A) 1.3 Ω B) 2.2 Ω C) 10 Ω D) 12 Ω

Rudiy27

Answer:

B) 2.2 Ω

Explanation:

First of all, we should notice that the resistor placed between A and B is short-circuited. In fact, the current from point A will follow the wire above between A and C (which has zero resistance), so we can basically ignore the presence of the resistor between A and B, since it has no effect on the circuit.

Then, we can notice that the two resistors between CB and CD are in parallel to each other; therefore, their equivalent resistance is given by:

$\frac{1}{R}=\frac{1}{R_{BC}}+\frac{1}{R_{CD}}=\frac{1}{4.0 \Omega}+\frac{1}{5.0 \Omega}=\frac{9}{20 \Omega}\\R= \frac{20 \Omega}{9}=2.2 \Omega$

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3 years ago

Can someone answer these?

Scorpion4ik [409]

Nine

Basically you are using d = r * t

Frequency is 1 / t

d is more commonly represented by lambda ( $\lambda$ ). Lambda is the wavelength.

so the formula $\lambda$ = c / f

More commonly it is written as $\lambda$ *f = c.

Givens

c = 3*10^8 m/s

$\lambda$ = 0.1 nm = 0.1 nm *[1 m /1 * 10^9 nm] = 0.1 * 10^ - 9 meters.

f = ??

f = $\dfrac{3*10^8}{0.1 * 10^-9}$

f = 3 * 10^8 * 10^9 / 0.1 = 3 * 10^18

Answer: C

Ten

$\lambda$ = c/f

$\lambda$ = 3*10^8 / 1 * 10^16

$\lambda$ = 3*10^-8 meters

$\lambda$ = 30 * 10^-9 meters = 30 nm

Answer: D

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3 years ago