The percentage error in the mass and speed are 2% and 3% respectively. What is the the maximum error in kinetic energy calculated using these quantities?
Answers
Answered by
45
Given :
Δm/m x 100= 2%
Δv/v x100= 3%
K.E=K= 1/2 mv²
(ΔK/K) x100=(Δm/m )x100+ 2 (Δv/v) x100
=2+2(3)
=2+6
=8%
∴The maximum error in kinetic energy IS 8%
Answered by
17
Hii mate,
# Answer- 8 %.
# Given-
Δm/m = 2 %
Δv/v = 3 %
# Explaination-
Let's solve this step-by-step.
Formula for kinetic energy is,
K.E. = (1/2)mv^2
Percentage error in kinetic energy will be,
ΔK.E./K.E.×100 = Δm/m×100 + 2Δv/v×100
Δd/d×100 = 2 % + 2×3 %
Δd/d×100 = 8 %.
Percentage error in kinetic energy is 8 %.
Hope that was useful...
# Answer- 8 %.
# Given-
Δm/m = 2 %
Δv/v = 3 %
# Explaination-
Let's solve this step-by-step.
Formula for kinetic energy is,
K.E. = (1/2)mv^2
Percentage error in kinetic energy will be,
ΔK.E./K.E.×100 = Δm/m×100 + 2Δv/v×100
Δd/d×100 = 2 % + 2×3 %
Δd/d×100 = 8 %.
Percentage error in kinetic energy is 8 %.
Hope that was useful...
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