I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
It’s not Cereal and Milk, carbonated drinks, and smoky air
Answer:
the density of gasoline is about 0.7 kg per latest a URL 50 litre of gasoline to your car and when feeling it what is the mass of the same amount of gasoline 2144 to 86 kg answer
you are so wise how do you do it?
Answer:
ε₂ =2.63 V
Explanation:
given,
M = 0.0034 H
I (t) = I₀ sin (ωt)
I (t) = 5.4 sin (143 t)
magnitude of the induced emf in the second coil
ε₂ =
ε₂ =
for maximum emf
cos (143 t) = 1
ε₂ =
ε₂ =2.63 V
