Answer:
Work done, W = 1.44 kJ
Explanation:
Given that,
Mass of boy, m = 74 kg
Initial speed of boy, u = 1.6 m/s
The boy then drops through a height of 1.56 m
Final speed of boy, v = 8.5 m/s
To find,
Non-conservative work was done on the boy.
Solution,
The work done by the non conservative forces is equal to the sum of total change in kinetic energy and total change in potential energy.



W = 1447.21 Joules
or
W = 1.44 kJ
Therefore, the non conservative work done on the boy is 1.44 kJ.
<u><em>21% of the atmosphere is Oxygen. </em></u>
<h2>Answer</h2>
1m/s
<h2>Explanation</h2>
Given that:
<em>Mass of first blob = 2kg = m1</em>
<em>Velocity of blob = 4m/s = v1</em>
<em>Mass of second blob = 6kg = m2</em>
<em>Velocity of blob = 0m/s = v2</em>
<em />
To find:
<em>Final velocity = Vf</em>
<em />
<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>
<em />
<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>
(2*4) + (6*0) = Vf(2+6)
8 + 0 = Vf(8)
8 = Vf(8)
Vf = 1 m/s
So the speed of two blobs immediately after colliding = 1 m/s
Answer:
Explanation:
#1 uses Newton's 2nd Law: F = ma so filling in:
750 = .4a and
a = 1875 m/s/s
#2 uses d = rt so
d = 12.3(25) and
d = 307.5 m