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Question Number 50274 by Tawa1 last updated on 15/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

10)∫_a ^b (Q/(2πε_0 ε_r r))dr  =(Q/(2πε_0 ε_r ))ln((b/a))  =((2×10^(−6) )/(2×3.14×8.85×10^(−12) ×2.77))ln2  =(1/(3.14×8.85×2.77))×10^6 ×ln2  =9004.77  pls check

10)abQ2πϵ0ϵrrdr=Q2πϵ0ϵrln(ba)=2×1062×3.14×8.85×1012×2.77ln2=13.14×8.85×2.77×106×ln2=9004.77plscheck

Commented by Tawa1 last updated on 15/Dec/18

God bless you sir

Godblessyousir

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

11)=(1/π)∫_0 ^π [10sinwt+3sin3wt+2sin5wt]d(wt)  =((−1)/π)∣10×((coswt)/1)+3×((cos3wt)/3)+2×((cos5wt)/5)∣_0 ^π   =((−1)/π)[10(cosπ−cos0)+(cos3π−cos0)+(2/5)(cos5π−cos0)]  =((−1)/π)[10(−1−1)+(−1−1)+(2/5)(−1−1)]  =((−1)/π)[(−2)(10+1+(2/5))]  =(2/π)×((57)/5)=((114)/(5π))  =7.26

11)=1π0π[10sinwt+3sin3wt+2sin5wt]d(wt)=1π10×coswt1+3×cos3wt3+2×cos5wt50π=1π[10(cosπcos0)+(cos3πcos0)+25(cos5πcos0)]=1π[10(11)+(11)+25(11)]=1π[(2)(10+1+25)]=2π×575=1145π=7.26

Commented by Tawa1 last updated on 15/Dec/18

God bless you sir, i appreciate your time sir

Godblessyousir,iappreciateyourtimesir

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