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Phantasy [73]
3 years ago
13

A length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

ing 17.0 ω. what was the resistance of the original length of wire before it was cut up? answer in units of ω.
Physics
1 answer:
ycow [4]3 years ago
6 0
The equivalent resistance of resistors connected in parallel is
\frac{1}{R_{eq}} = \frac{1}{R_1} +  \frac{1}{R_2} + ...
In our problem, we have 5 identical resistor of resistance R, so their equivalent resistance is
\frac{1}{R_{eq}} =  5 (\frac{1}{R} )
The problem also says that the equivalent resistance is R_{eq}= 17 \Omega, so we can find the resistance R of each piece of wire:
R=5 R_{eq} = 5 \cdot 17 \Omega = 85 \Omega

In the initial wire, it's like the 5 pieces are connected in series, so the equivalent resistance of the wire is just the sum of the resistances of the 5 pieces:
R_{wire} = 5 R = 5 \cdot 85 \Omega = 425 \Omega
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y-component of R = 12 + 14 - 6 = 20 N

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y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

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