Recent content by Keshroom

  1. K

    How does one get the form of a circle out of this equation?

    ofcause!! oh how foolish i am :) answer: 1/9 = (x+ 5/3)^2 + y^2 thanks guys
  2. K

    How does one get the form of a circle out of this equation?

    Homework Statement Sketch modulus((z+1)/(2z+3))=1 on the complex plane where z=x+iy Homework Equations The Attempt at a Solution I know it is a circle but i need help simplifying the equation into the form of a circle. i'm stuck at 0= 3x^2 + 3y^2 + 10x + 8 I usually...
  3. K

    How to count the number of occurences of an integer in excel?

    Homework Statement I have a thousand random numbers generated between -100 and 100 and want to count the number of occurences of each integer. I know i can use the COUNTIF function, but for so many numbers it takes way too long. Is there a way to create a loop in excel that can do this or...
  4. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    alright i understand the theory behind it. But assume we didn't know anything about how this question is related to kinetic energy, how would i begin to solve it? Since it was a question in my math class and you are not expected to know kinetic energy.
  5. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    ok, so if T' is zero, that means acceleration =0 so therefore the speed will be constant. right?
  6. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    dT/dt = mv.dv/dt when m =1 dT/dt= v.dv/dt i can simplify furthur, but why? v' is the derivative of v ( velocity )
  7. K

    Show that if v . v' = 0 (both vectors) then speed v is constant

    Homework Statement Show that for a particle moving with velocity v(t), if v . v'=0 then the speed v is constant. Homework Equations The Attempt at a Solution Let v = (v1,v2,...,vn) and let v'= (v'1,v'2,...,v'n) So, v . v'= v1v'1 + v2v'2 + ... + vnv'n = 0 This is the...
  8. K

    Finding x(t) from a time dependant force

    haha yeah stop doing my homework for me! Ok i figured it out properly F(t) = ma0e-bt m(dv/dt) = ma0e-bt ∫mdv = ∫ma0e-btdt (from v=0 to v=v and t=0 to t=t) mv = m(a0e-bt)/-b (dx/dt) = (a0e-bt)/-b ∫dx = ∫(a0e-bt)/-b dt (from x=0 to x=x and t=0 to t=t) x(t) = a0e-bt/b2
  9. K

    Finding x(t) from a time dependant force

    alright ill take your word for it!
  10. K

    Finding x(t) from a time dependant force

    wait..you can just integrate like that without the dt on the right hand side?
  11. K

    Finding x(t) from a time dependant force

    touche sir. That does not make sense. ahh i see now Question was F(t) = ma0e-bt now i know why it is a0 so in this case acceleration a = a0e-bt but i'm still stuck! How do i get to x(t)?
  12. K

    Finding x(t) from a time dependant force

    yeah... this is what i tried F(t) = m(dv/dt) e-bt F(t)ebtdt = mdv ∫F(t)ebtdt = ∫mdv (with bounds t=0 to t=t, and v=0 to v=v) ∫F(t)ebtdt = mv (only integrating right side yields) ∫F(t)ebtdt = m(dx/dt) [∫F(t)ebtdt]dt = dx (integrating both...
  13. K

    Finding x(t) from a time dependant force

    Homework Statement A particle of mass m is subject to a force F(t) = mae-bt The initial position and speed are zero. Find x(t) Homework Equations The Attempt at a Solution how would i begin to start this??
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