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  1. Part 1 Find time period (T) from the formula T = 2*pi*(M/k)^0.5 Then use T = 2*pi / ! Part 2 At equilibrium v = A* ! From here calculate A but this A is w.r.t...

    2 Answers · Science & Mathematics · 17/07/2011

  2. ...and ω = √(k/m) V(max) = A√(k/m) V(max) / A = √(k/m) (V(max) / A)² = k/m k = (V(max) / A)²m k = (100 / 0.10)² * 5 k = 5 000 000...

    1 Answers · Science & Mathematics · 11/06/2017

  3. ... - Be Mine : http://www.youtube.com/watch?v=amvKLtj6jVE Some other good boy ...dance to are : U-KISS, EXO (M & K), NU'EST, BtoB, SHINee...

    3 Answers · Arts & Humanities · 27/05/2012

  4. ...hexadecimal for any positive decimal number *\ #include<stdio.h> void main() { int a[32],n,c,i,m,k,v; printf("enter ur number"); scanf("%d",&n); for(i=0;i<3;i++) { if (i==0) c=2...

    3 Answers · Computers & Internet · 18/10/2007

  5. 1) V = A*√[k/m] 2) a = A*[k/m] Divide (1) by (2) to get V/a = √[m/k] Use this value in the formula for T: T = 2π√[m/k]

    1 Answers · Science & Mathematics · 09/12/2011

  6. ... directly proportional to d(mv)/dt. F = k d(mv)/dt F = m k d(v)/dt F = m k a k is just a proportionality constant so that the units of force, mass...

    2 Answers · Science & Mathematics · 11/09/2012

  7. ... a=(k^2)r(t^2) using F=ma F/m =(k^2)r(t^2) F=m(k^2)r(t^2) E= F x...m are constant so; 1/v is proportional to a v =s/t 1/v =t/s t/s proportional to a Substitute...

    1 Answers · Science & Mathematics · 29/03/2010

  8. ...^2 where V is max v. We solve for V and get... V = ± A √(k/m) = ± √( (k*x0^2+m*v0^2)/m )

    1 Answers · Science & Mathematics · 30/04/2009

  9. ... and C are increased m times. a new = k m^2 AB and v new = kk’m^3 ABC [v/a] new = m k’C = m [v/a] initial. Hence the proof. ================================

    3 Answers · Science & Mathematics · 16/02/2008

  10. to find the time t=2(pi)√(m/k) t=0.3410s amplitudde v=ω(√(a^2-y^2)) v(max)=aω ω=√(k/m) a=v(max)*√(m/k) a=0.01085 m acc(max) acc=a*ω^2 acc=a*v(k/m)^2 acc=a*(k/m) acc=3.69ms^2 hope this helps:D

    2 Answers · Science & Mathematics · 22/11/2011

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