Answer:
It relaxes
Explanation:
I took the test I hope it helps
Astronomers can measure a star's position once, and then again 6 months later and calculate the apparent change in position. The star's apparent motion is called stellar parallax. The distance d is measured in parsecs and the parallax angle p is measured in arcseconds.
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The Correct choice is ~
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Answer:
34.4Joules
Explanation:
Complete question
a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal. Find the workdone
Work done = Fdsin theta
Force F = 20N
distance d = 3m
theta = 35 degrees
Substitute
Workdone = 20(3)sin 35
Workdone = 60sin35
Workdone = 34.4Joules
Hence the workdone by the man is 34.4Joules
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)
