1) In any collision the momentum is conserved
(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')
candel all the m factors (because they appear in all the terms on both sides of the equation)
2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')
2) Elastic collision => conservation of energy
=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2
cancel all the 1/2 and m factors =>
2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>
4(vo)^2 = 2(v1')^2 + (v2')^2
now replace (v2') = -2(v1')
=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>
(v1')^2 = [4/6] (vo)^2 =>
(v1')^2 = [2/3] (vo)^2 =>
(v1') = [√(2/3)]*(vo)
Answer: (v1') = [√(2/3)]*(vo)
When action potentials reach the end of the axon, they stimulate opening of Ca2+ channels, causing a release of neurotransmitters to the post-synaptic cell. How does and impulse propagate down the axon? The stimulus causes a start of the action potential and it moves down the axon without the ions moving down
The unit for measuring current is an " ammeter ".
The unit for measuring voltage is a " voltmeter ".
The unit of measure for electrical current is the " Ampere ".
The unit of measure for potential difference (voltage) is the " volt ".
So you would divide 1530 by 8 and that’s how you’d get your answer, so it should be (blank)m
