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Rzqust [24]
3 years ago
6

What elements are good conductors of electricity?

Physics
1 answer:
kramer3 years ago
6 0
In solids: All metals are good conductors of electricity as they contain free moving electrons. Non-metals doesn't conduct , but we consider Graphite the only non-metal that can conduct electricity for the presence of free moving electrons.

In Liquids ; Ionic compunds contains free moving ions , so they conduct electricity as well .

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42 I THE CORRECT ANSWER

Explanation:

BECAUSE I KNOW

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What is the acceleration of the car at segment C?
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-30 I think sorry if I’m wrong
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An iron sphere of radius 0.18m has mass 190 kg. Calculate the density of the iron
Slav-nsk [51]
  • radius=r=0.18m

Volume:-

\\ \sf\rightarrowtail \dfrac{4}{3}\pi r^3

\\ \sf\rightarrowtail \dfrac{4}{3}\pi (0.18)^3

\\ \sf\rightarrowtail \dfrac{4}{3}\pi (0.005832)

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8 0
2 years ago
A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows
jeka57 [31]

Answer:

(a) The impedance in the circuit is Z=183.33\Omega.

(b)The resistance is R=38.89\Omega.

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

Z=\frac{V_rms}{I_rms}

Here, V_rms is the rms voltage and I_rms is the rms current.

PutV_rms=110 V and I_rms=0.600 A.

Z=\frac{110}{0.600}

Z=183.33\Omega

Therefore, the impedance in the circuit is Z=183.33\Omega.

(b)

The expression for the average power is as follows;

P_{a}=I_{rms}^{2}R

Here, P_{a} is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

R=\frac{P_{a}}{I_{rms}^{2}}

Put P_{a}=14W and

R=\frac{14}{{0.600}^{2}}

R=38.89\Omega

Therefore, the resistance is R=38.89\Omega.

(c)

The expression for the impedance is as follows;

Z^{2}=R^{2}+X_{L}^{2}

Here,X_{L} is the inductive reactance.

Put Z=183.33\Omega and R=38.89\Omega.

(183.33)^{2}=(38.89)^{2}+X_{L}^{2}

X_{L}=179.16\Omega

The expression for the inductive reactance in terms of  frequency is as follows;

X_{L}=2\pi fL

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

L=\frac{X_{L}}{2\pi f}

Put X_{L}=179.16\Omega and f=50Hz.

L=\frac{179.16}{2\pi (50)}

L=0.57 H

Therefore, the inuctance is 0.57 H.

4 0
3 years ago
What do you mean by 100W is written on an electric bulb? ​
Goryan [66]

Answer:

100 watt

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