answer fast..... it's urgent
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E(t)=A2 exp(−αt)E(t)=A2 exp(−αt)
δ(E)=2A(δA) exp (−αt)+A2δ exp(−αt)δ(E)=2A(δA) exp (−αt)+A2δ exp(−αt)
⇒δ(E)=2A exp (−δt) .δA+A2 exp (−αt) δ(αt)⇒δ(E)=2A exp (−δt) .δA+A2 exp (−αt) δ(αt)
⇒δ(E)=2A exp (−αt) δA+A2 exp (−αt)(αδt+ t δα)⇒δ(E)=2A exp (−αt) δA+A2 exp (−αt)(αδt+ t δα)
⇒(δE)E=2(δAA)+α(δt)=2×0.0125+0.2×0.015×5⇒(δE)E=2(δAA)+α(δt)=2×0.0125+0.2×0.015×5
⇒δ(E)E=0.0225+0.015=0.040⇒δ(E)E=0.0225+0.015=0.040
% error = 4%
δ(E)=2A(δA) exp (−αt)+A2δ exp(−αt)δ(E)=2A(δA) exp (−αt)+A2δ exp(−αt)
⇒δ(E)=2A exp (−δt) .δA+A2 exp (−αt) δ(αt)⇒δ(E)=2A exp (−δt) .δA+A2 exp (−αt) δ(αt)
⇒δ(E)=2A exp (−αt) δA+A2 exp (−αt)(αδt+ t δα)⇒δ(E)=2A exp (−αt) δA+A2 exp (−αt)(αδt+ t δα)
⇒(δE)E=2(δAA)+α(δt)=2×0.0125+0.2×0.015×5⇒(δE)E=2(δAA)+α(δt)=2×0.0125+0.2×0.015×5
⇒δ(E)E=0.0225+0.015=0.040⇒δ(E)E=0.0225+0.015=0.040
% error = 4%
jayakulpreet:
bt not preparing for anything
I also have my own solution bt trying to find out the easy one
do u want to have that solution
no i have one
more easy than yours
ohk
we r of same age
yes
ok great anything else
no
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