The energy of a system as a function of time tis
given as E (t) = Aexp (-at), where a = 0-2 s-!
The measurement of A has an error of 1.25%. If
the error in the measurement of time is 1.50%, the
percentage error is the value of E (t) at t = 5 sec is
Answers
Answered by
1
Answer:
E(t)=A
2
e
−αt
Taking natural logarithm on both sides,
ln(E)=2ln(A)+(−αt)
Differentiating both sides
E
dE
=2
A
dA
+(αdt)
Errors always add up for maximum error.
∴
E
dE
=2
A
dA
+α(
t
dt
)×t
Here,
A
dA
=1.25 %,
t
dt
=1.5%, t=5s, α=0.2s
−1
∴
E
dE
=(2×1.25%)+(0.2)×(1.5%)×5=4%
I think this is it.
Answered by
0
ANSWER
E(t)=A2e−αt
Taking natural logarithm on both sides,
ln(E)=2ln(A)+(−αt)
Differentiating both sides
EdE=2AdA+(αdt)
Errors always add up for maximum error.
∴EdE=2AdA+α(tdt)×t
Here, AdA=1.25 %, tdt=1.5%, t=5s, α=0.2s−1
∴EdE=(2×1.25%)+(0.2)×(1.5%)×5=4%
Answer By Jannatk0218
E(t)=A2e−αt
Taking natural logarithm on both sides,
ln(E)=2ln(A)+(−αt)
Differentiating both sides
EdE=2AdA+(αdt)
Errors always add up for maximum error.
∴EdE=2AdA+α(tdt)×t
Here, AdA=1.25 %, tdt=1.5%, t=5s, α=0.2s−1
∴EdE=(2×1.25%)+(0.2)×(1.5%)×5=4%
Answer By Jannatk0218
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