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  1. If constant acceleration, average speed is the half the final speed. Average speed = (total distance traveled)/(time spent on a journey)

    1 Answers · Science & Mathematics · 12/05/2011

  2. ..., if the car is traveling at a constant speed, that means acceleration is zero, which means... zero, that means the frictional force would need to be exactly opposite to whatever other forces were...

    1 Answers · Science & Mathematics · 09/11/2009

  3. ...force to the gravitational force: mv^2/r = GMm/r^2 We get the equation for the speed required to keep the stone in orbit: v = √GM/r Input the...

    1 Answers · Science & Mathematics · 01/02/2011

  4. ...loop so that it doesn't fall. The increase in speed is there to compensate for frictional and potential energy losses as it...to the top. Technically, the car doesn't need to increase it's speed if it is going...

    2 Answers · Science & Mathematics · 07/04/2011

  5. If you have two of these, velocity(V) , Distance (S) and Time (T) You can find the third one with this formula V = S/T

    4 Answers · Science & Mathematics · 12/10/2011

  6. You need the linear density of the string: v = √[F/μ] μ = F₁/v₁²...

    5 Answers · Science & Mathematics · 01/05/2010

  7. ½mv² = mK(T-30) where K is the specific heat of lead and T is the melting temperature in °C Solve this eq for v

    1 Answers · Science & Mathematics · 15/04/2008

  8. ...g) sin 2 θ without using too much complicated math we can see that for any given Vo the max R will be when sin 2 θ is also...

    1 Answers · Science & Mathematics · 27/05/2015

  9. ... at θ = 45º where sin2θ = 1 and is d = v0²/g For a distance of 3187 ft, v0 = √[3187*32.17] = 320 ft/s This is 218...

    3 Answers · Science & Mathematics · 17/03/2009

  10. There are five constant acceleration formulae, each one omits one of the variables. I suggest you learn them off by heart. v = u + at v^2 = u^2 + 2as s = ut + 0.5at^2 s = vt - 0.5at^2 s = (u + v)t / 2 u = initial velocity (m/s) v = final velocity (m/s) s = distance or...

    4 Answers · Science & Mathematics · 03/10/2009

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