A 1200-kg car initially at rest undergoes constant acceleration for 8.8 s, reaching a speed of 10 m/ s. It then collides with a
1 answer:
Answer:
The final kinetic energy of the two-car system is 60,000 J.
Explanation:
Given;
mass of the car, m = 1200 kg
time of motion, t = 8.8 s
final velocity of the car, v = 10 m/s
Apply the principle of conservation of kinetic energy; the initial kinetic energy is equal final kinetic energy.
Therefore, the final kinetic energy of the two-car system is 60,000 J.
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Answer:
Hello the diagram related to your question is attached below
answer: a) 851 m/s
b) 8506.1 secs
Explanation:
calculate the periodic time of the satellite using the equation below
t = -- ( 1 )
where ; R = 6370 km
h = 500 km
g = 9.81 m/s^2
input given values into equation 1
t = 5670.75 secs
next calculate the periodic time taken by the space craft
<u>a) determine the increase in speed </u>
V = v -
where ; v = 8463 m/s , R = 6370 km, h = 500 km
V = 851 m/s
b) Determine the periodic time for the elliptic orbit
τ =
= = 8506.1 secs
attached below is the remaining part of the detailed solution
Answer:
Her speed at the bottom of the slope is 25.665 m/s
Explanation:
Here we have
Initial velocity, v₁= 15 m/s
Final velocity = v₂
The energy balance present in the system can be represented as
Where:
m = Mass of the cyclist = 70 kg
W = work done by the drag force =
Where:
d = Distance traveled = 450 m
Therefore,
and
= 658.714 m²/s²
v₂ = 25.665 m/s
Her speed at the bottom of the slope = 25.665 m/s.