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(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

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

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

Q = m * Cp * dT Heat = mass * Specific Heat * Temperature Change 1635J = 2000g * Cp * 3K Cp = 1635/(2000 * 3) 0.2725 J/(g*K) or 272.5 J/(kg*K)

1 Answers · Science & Mathematics · 28/09/2008

height=initial velocity times time --1/2gravity times time squared h=2t-4.905t^2 you need to find the time when the velocity is zero 0=v-at t=.204 seconds plug it back in to the first equation h=2(.204)-4.905(.204)^2 h=.204 meters

1 Answers · Science & Mathematics · 04/10/2011

g = - 10 (It's going down. vi = 20 m/s t = 1 second vf = ???? g = (vf - vi)/t -10 = (vf - 20)/1 -10 = vf - 20 10 = vf This means it is still going up and the speed is 10 m/s up -10 = (vf - vi)/2 -20 = vf - 20 vf = 0 I leave the rest to you...

1 Answers · Science & Mathematics · 29/08/2011

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

(2m/s) / (9.81m/s^2) = 0.204s He will start to fall down after 0.204 seconds, this is the moment where he will be the highest. d=-(a/2)t^2+vt d is how much the center of mass is raised, a is acceleration, v is initial velocity, t is time d=-(9.81/2)t^2...

1 Answers · Science & Mathematics · 07/12/2010

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

Momentum is conserved; the initial momentum of the system is zero, as is the final momentum: m*V = M*v → v = V*m/M = 4.1*15/188 v = 0.327 m/s

1 Answers · Science & Mathematics · 03/12/2012

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

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