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>
Answer:
Sounds cool.. but what do they do?
Explanation:
Oxygen is like a milder form of nitrogen when it comes to its changing forms.
Oxygen becomes a gas when heated above -183 degrees Celsius, and when it is cooled below -183 degrees Celsius, it becomes a liquid.
Just as a bonus, the transition between liquid and solid forms happens when Oxygen is cooled below -218.79 degrees Celsius.
So your answer is "the conversion between the liquid and gaseous states of Oxygen." :)
The electrons in the positive object are attracted to the negatively charged object. Some of the electrons move from the positive object to the negative object.
an example of this is lightning because the positive electrons on/in the earth are attracted to the negative electrons in the clouds and sky so the positive move to the negative charge.
