Let original length be L. The new length is therefore 4L.
Let original cross sectional surface area of the wire be equal to πr^2.
This means original volume was L x πr^2 = Lπr^2
The volume is the same but the length is different so 4L x new surface area must be equal to Lπr^2. Let new surface area be equal to Y.
4L x Y = Lπr^2
=> Y = (πr^2 )/ 4
Using the resistivity formula,
R = pL/A. p which is resistivity is a constant so it stays the same
But this time, instead of L we have 4L and instead of πr^2 we have (πr^2)/4.
so the new resistance
= (4Lp)/ {(πr^2)/4}
= 16 (pL)/(πr^2)
= 16 (pL)/A. because πr^2 is A
since pL/A is equal to R from the formula, this is equal to
16 R.
R was 10 ohms
therefore new resistance is 16 x 10 = 160 ohms
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Answer:
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Explanation:
Answer:
100m
Explanation:
100m
s=ut+1/2at^2
s= unknown, u=0, a=2, t=10
s=0*10+1/2(2)(10)^2
s=1/2(2)(100)
s=1(100)
displacement = 100 meters
C2 = a2 + b2
29^2 = a^2 + 20^2
841 = a^2 + 400
441 = a^2
a = 21
