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Snezhnost [94]
3 years ago
7

If the voltage is 120 v, and the desired current is 5 a, what resistance must be in the circuit?

Physics
1 answer:
ch4aika [34]3 years ago
7 0
Resistance = (voltage) / (current)

= 24 ohms.
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olga55 [171]

Answer:

The value is V  =  2 V

Explanation:

From the question we are told that

The length of the wire is l = 10 \ m

The current density is J =  4*10^6 \  A/m^2

The conductivity is \sigma  =  2*10^{7} \  S/m

Generally conductivity is mathematically represented as

\sigma  =  \frac{l}{RA}

Here R is the resistance which is mathematically represented as

R =  \frac{V}{I}

Here I is the current which is mathematically represented as

I  =  J * A

So

R =  \frac{V}{  J * A}

And

\sigma  =  \frac{l}{\frac{V}{  J * A} * A}

=> \sigma  =  \frac{l}{\frac{V}{J}}

=> V = \frac{l * J}{\sigma }

=> V = \frac{10  * 4*10^6}{2*10^{7}  }

=> V  =  2 V

5 0
3 years ago
An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t
scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

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Now we can isolate V, the potential difference between the plates as:

V = \frac{Q}{e_0} *\frac{d}{A}

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V

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The question is on the picture
Firlakuza [10]

Answer:

<em> think 2 also if not im so sorry  but i think it is :)</em>

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For 40 hours, she gets paid $340
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