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)
Answer:
Output power = 500 KW
Explanation:
Given the following data;
Efficiency = 20%
Input power = 2500 KW
To find the output power;
Substituting into the equation, we have;
LET Output power = OP
Cross-multiplying, we have;
Output power = 500 KW
I believe it is speed.
Hope this helps!
D) the north star is not part of our solar system...
hope this helps!
We need to find the average speed of the ball during the motion of 1 m
In order to find that we took several reading and found following times to cover the distance of 1 m
t1 = 2.26 s
t2 = 2.38 s
t3 = 3.02 s
t4 = 2.26 s
t5 = 2.31 s
Now in order to find the average time we can write
So average time to cover the distance of 1 m by ball will be 2.45 s
here 3.02 s is not the average time but we can say it is the median of the readings of all possible values which we can not use in our calculation as average time