<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
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Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
Answer:
It's energy will double.
Explanation:
This is because energy, E, is related to frequency, f, by:
E = hf
Where h = Planck's constant
So, double frequency will be 2f
=> E(2f) = 2hf = 2E.
Hence, energy is doubled.
