If the percentage error in the measurement of velocity and mass are 2% and 4% respectively. what is the percentage error in k.e
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given,
∆v/v×100=2%
∆m/m ×100=4%
we know that ,
k.e = 1/2mv^2
then
∆k.e/k.e×100=∆m/m×100 + 2(∆v/v×100)
=4+ 2×2
=4+4
=8%
Hence, the percentage error in k.e is 8%
∆v/v×100=2%
∆m/m ×100=4%
we know that ,
k.e = 1/2mv^2
then
∆k.e/k.e×100=∆m/m×100 + 2(∆v/v×100)
=4+ 2×2
=4+4
=8%
Hence, the percentage error in k.e is 8%
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