The tension developed in the spring by stretching 2cm is 20N. Then spring constant of spring is
1) 1000Nm^-1
2) 500Nm^-1
3) 1500Nm^-1
4) 100Nm^-1
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Answer:
third number 1500 INR in UPA 1
Answered by
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Answer:
The answer is 1000N/m
Explanation:
T= -k x
20= -k 2×10^-2
2000 = - 2k
then; k = 1000 N/m
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