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2 years ago

A(n) _____ is an electric circuit that produces a magnetic field.

Physics

2 answers:

ArbitrLikvidat [17]2 years ago

7 0

The answer is electromagnet

netineya [11]2 years ago

4 0

Magnetic Battery

This is a device which converts magnetic field energy into electrical energy so in this part we got the electric energy out of it

Coil Magnet

This type of magnet is always in shape of coil which will produce the magnetic field along the axis.

Coil circuit

This is just type of the circuit which will produce the current in closed loop.

Electromagnet

In electromagnets we got the magnetic field energy which is converted by electric energy i.e. in this form we have to supply the current in the circuit and then it will produce the magnetic field of that current

So correct answer is electromagnet

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n general, which trophic level has the LEAST energy available to it? A) producer B) primary consumer C) secondary consumer D) te

zaharov [31]

D - tertiary consumer

This is because it is the farther up to food chain.

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2 years ago

An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-f

IgorC [24]

Resistance = (voltage) / (current)

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2 years ago

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

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2 years ago

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2 years ago

Read 2 more answers

Two small nonconducting spheres have a total charge of 93.0 μC . Part A

Travka [436]

Answer:

charge on each

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

Q1Q2 = 1.49 × $10^{-9}$ C²

and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

5 0

2 years ago