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  1. Your picture did not open. I am trying to answer your questions as far as possible with the help of your word problem only. 1) Use energy conservation here. Net change in kinetic energy = 0.5 mv^2 [as final speed = 0] = 4.86 J This...

    1 Answers · Science & Mathematics · 19/01/2014

  2. that... doesn't seem to be a valid question. First, there's no units, so it's impossible to answer it for that reason alone. Second, if you want units of work, what you need is force and distance, not speed. W = FD And power is Work per unit time so...

    1 Answers · Science & Mathematics · 15/01/2009

  3. You need to use terminal velocity ideas. Check this link: http://en.wikipedia.org/wiki/Stokes%27_law Hope this helps. your_guide123@yahoo.com

    1 Answers · Science & Mathematics · 18/11/2012

  4. Acceleration = (f/m) = 4,000/2,000) = 2m/sec^2. Maximum V = sqrt.(ar) = 20m/sec. (20 x 3.6) = 72kph., should you need that.

    1 Answers · Science & Mathematics · 02/02/2014

  5. m1 v1 + m2 v2 = m1 u1 + m2 u2 m1 = 69 + 133 = 202 kg m2 = 15 kg v1 = v2 = 0 u2 - u1 = 4.4 m/s u2 = u1 + 4.4 m1 v1 + m2 v2 = m1 u1 + m2 u2 0 = 202 u1 + 15 (u1 + 4.4) 0 = 202 u1+ 15 u1 + 66 -66 = 217 u1 u1 = -66/217 = -0.304 m/s (it negative because the direction is opposed the package direction)

    1 Answers · Science & Mathematics · 14/03/2013

  6. a) 0.5*m*v^2 = 12 J b) F*d = 600 J c) F*d = -200J d) F*d = m*g*(8sin(37)) = -283.38 J e) F*d = 0 J f) W_net = 600 - 200 - 283.38 = 116.62 J

    1 Answers · Science & Mathematics · 29/11/2009

  7. When you driving at 45,6 m/s and you reduces your speed uniformly to 25.0 m/s you drive on the average 35.3 m/s In 4.20 s that is 148.26 m. So can also use the formal equations: a = (v1 - v0) / t. = (45.6 - 35.3)/4.2 and s = v0.t + (1/2).a.t^2 with s (at t=0) = 0...

    1 Answers · Science & Mathematics · 06/10/2010

  8. (a) p = mv = 1.2e5 kg·m/s (b) v = p / m = 1.2e5kg·m/s / 8kg = 1.5e4 m/s = 15 km/s

    1 Answers · Science & Mathematics · 14/10/2013

  9. By I = ∆P = m∆v =>9.87 = 0.54 x ∆v =>∆v = 18.28 m/s

    1 Answers · Science & Mathematics · 21/10/2013

  10. ENERGY AT THE TOP = ENERGY AT THE BOTTOM, FIND THE SPEED AT THE BOTTOM 1/2*m*v1^2 + m*g*h1 = 1/2*m*v2^2 + m*g*h2 In this case, set h2 = 0. All the m's cancel 1/2*v1^2 + g*h1...

    1 Answers · Science & Mathematics · 09/12/2012

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