If ALL of the spring energy becomes KE, then 400 J = ½mv² = ½ * 0.2kg * v² → → v = 63 m/s That's a high speed, but it's also a lot of work for a rubber band.

1 Answers · Science & Mathematics · 25/03/2021

KE becomes PE means that ½mv² = mgh. Rearrange to h = v² / 2g Plug in v and g.

1 Answers · Science & Mathematics · 26/03/2021

...from the ramp base is R = 7.16 + 1.31 = 8.47 m if fired and

**released**at the same instant. Telling us that the block starts from 10 m...1 Answers · Science & Mathematics · 09/04/2021

E = hf where h is a constant called Planck's Constant. E is the energy

**released**in the transition = Ei - Ef and f is the frequency of the emitted photon.1 Answers · Science & Mathematics · 04/04/2021

...) = 42,525 J So a huge amount of internal energy was

**released**in this collision. Where did that energy come from? No solid ...1 Answers · Science & Mathematics · 22/03/2021

... = v - w .... where w is the final velocity of right block just after spring release 3v/2 = v - w w = v + 3v/2 w = 5v/2 The right block will have velocity of 5v...

1 Answers · Science & Mathematics · 26/03/2021

(a) EPE = ½kx² = ½ * 434N/m * (0.375m)² = 30.5 J (b) KE = 30.5 J = ½mv² = ½ * 0.316kg * V² → → V = 13.9 m/s

2 Answers · Science & Mathematics · 17/03/2021

a) for an object on a frictionless incline, the acceleration is a = g*sinΘ = 9.8m/s² * sin12º = 2 m/s² Then using s = ½at² we get t = √(2s / a) = √(2 * 6m / 2m/s²) = 2.4 s b) Now the acceleration is a = g*sinΘ...

2 Answers · Science & Mathematics · 25/03/2021

(a) Since you have a nice "round" number like 576 ft, I think you want to use a nice "round" number for g. Then h(t) = h₀ + v₀*t + ½at² = 576ft + 0*t - ½*32ft/s²*t² h(t) = 576 - 16t² ◄ for...

3 Answers · Science & Mathematics · 04/04/2021

initial separation is d = √(1.25² + 0.570²) m = 1.374 m initial energy is E = kQq / d = 8.99e9N·m²/C² * 2.72e-6C * 3.30e-6C / 1.374m E = 0.05874 J At "half the speed it will have at infinity," the KE...

1 Answers · Science & Mathematics · 18/03/2021