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1. The resistance of a wire is 20 ohm. It is so stretched that the length becomes three times, then the new resistance of the wire is?
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Answers
Answered by
3
R=(ρ∗L)/AR=(ρ∗L)/A
If the wire is an isolated system (no extra metals/wires coming in or going out) with a specified length and cross-sectional area, then stretching it would increase the length 3 times and presumably reduce the cross-sectional area 3 times as well. Then,
R=(ρ∗3L)/(1/3∗A)R=(ρ∗3L)/(1/3∗A)
R=9(ρ∗L)/A)R=9(ρ∗L)/A)
You get 9 times more resistance you initially had, this makes sense as you bring the wire to a thinner cross-sectional area, the electrons have more material to go through.
Thanks
hope you like it
please mark as a brainliest answer
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If the wire is an isolated system (no extra metals/wires coming in or going out) with a specified length and cross-sectional area, then stretching it would increase the length 3 times and presumably reduce the cross-sectional area 3 times as well. Then,
R=(ρ∗3L)/(1/3∗A)R=(ρ∗3L)/(1/3∗A)
R=9(ρ∗L)/A)R=9(ρ∗L)/A)
You get 9 times more resistance you initially had, this makes sense as you bring the wire to a thinner cross-sectional area, the electrons have more material to go through.
Thanks
hope you like it
please mark as a brainliest answer
^_^
Juliet789:
hi
where??
where down??
Answered by
1
resistance= ro l/A
where ro is resistivity constant, l is length of wire and A is cross-sectional area
l2=3times of l1
since, l is directly proportional to resistance,
resistance is increased threefold
therefore resistance is 3 x 20 ohm = 60ihm
Hope this helps :)
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