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3 years ago

7

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

running through the entire circuit?

1 answer:

larisa [96]3 years ago

6 0

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega$

And now we can use Ohm's law to find the current in the circuit:

$V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A$

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0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)

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We want to find the values of x, y, z such that the above equation is true.

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Now in order to find the average time we can write

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=> $T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }$

=> $T = 53473 \ second = 14.8 \ hours$

Answer:

The orbital velocity $v = 16.4 \ km/s$

The period is $T = 14.8 \ hours$

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as

$\frac{GM_s * m }{r^2 } = m w^2 r$

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