There is no diagram below so I can't answer the question
Explanation:
Both graphs show plotted points forming a curved line. Curved lines have changing slope; they may start with a very small slope and begin curving sharply (either upwards or downwards) towards a large slope. In either case, the curved line of changing slope is a sign of accelerated motion (i.e., changing velocity).
To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as
As M=m, then
Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then
Therefore replacing we have that,
Re-arrange to find v,
Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that
Therefore
Therefore the velocity when they are about to collide is 
Answer:
0.96 m
Explanation:
First, convert km/h to m/s.
162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s
Now find the time it takes to move 20 m horizontally.
Δx = v₀ t + ½ at²
20 m = (45.08 m/s) t + ½ (0 m/s²) t²
t = 0.4436 s
Finally, find how far the ball falls in that time.
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²
Δy = -0.96 m
The ball will have fallen 0.96 meters.
<h3>
Answer: 104.5 cubic cm</h3>
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Work Shown:
r = radius = 1.045 cm
h = height = 30.48 cm
pi = 3.141 approximately
V = volume of cylinder
V = pi*r^2*h
V = 3.141*(1.045)^2*30.48
V = 104.547940002
V = 104.5 cubic cm
