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Question Number 50835 by Tawa1 last updated on 21/Dec/18

Commented by Tawa1 last updated on 21/Dec/18

Please help me solve it with diagram

Pleasehelpmesolveitwithdiagram

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18

1)i)acc=w^2 x=4π^2 n^2 x=4×π^2 ×50^2 ×2×10^(−2)      =200π^2 =1973.92meter/sec^2   velocity v=w(√(a^2 −x^2 ))   velocity at middle v=2πn(√(a^2 −x^2 ))   v=2π×50×(√((4×10^(−2) )^2 −(2×10^(−2) )^2 ))   =100π×10^(−2) (√(12))   =π(√(12)) metrt/sec  ii)acc=4π^2 n^2 x  =4π^2 ×50^2 ×4×10^(−2) =400π^2 =3947.84m/sec^2   velocity at 4cm=0  pls che4k...

1)i)acc=w2x=4π2n2x=4×π2×502×2×102=200π2=1973.92meter/sec2velocityv=wa2x2velocityatmiddlev=2πna2x2v=2π×50×(4×102)2(2×102)2=100π×10212=π12metrt/secii)acc=4π2n2x=4π2×502×4×102=400π2=3947.84m/sec2velocityat4cm=0plsche4k...

Commented by Tawa1 last updated on 21/Dec/18

God bless you sir. Any diagram to it sir and the second one

Godblessyousir.Anydiagramtoitsirandthesecondone

Commented by Tawa1 last updated on 21/Dec/18

God bless you sir. Any diagram sir with the second question

Godblessyousir.Anydiagramsirwiththesecondquestion

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Dec/18

is the answer correct for question no1...  i am trying to solve 2nd questiin...

istheanswercorrectforquestionno1...iamtryingtosolve2ndquestiin...

Commented by Tawa1 last updated on 21/Dec/18

i don′t know sir but i will use your workings

idontknowsirbutiwilluseyourworkings

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