A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east.
1 answer:
Answer:
Time, t = 12 minutes
Explanation:
It is given that,
A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :
d = 4800 m
Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s
Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :
t = 719.99 seconds
t = 720 seconds
or
t = 12 minutes
So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.
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however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
and the sign is negative because the frictional force acts against the direction of motion of the dragster.
This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is
:
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And since 1 horsepower is equal to 746 W, we can rewrite the power as
however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
and the sign is negative because the frictional force acts against the direction of motion of the dragster.
This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is
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Answer:
The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s
Explanation:
The rotational inertia of the merry-go-round = 600 kg·m²
The radius of the merry-go-round = 3.0 m
The mass of the boy = 20 kg
The speed with which the boy approaches the merry-go-round = 5.0 m/s
Where;
= The tangential force
I = The rotational inertia
m = The mass
α = The angular acceleration
r = The radius of the merry-go-round
For the merry go round, we have;
= The rotational inertia of the merry-go-round
= The angular acceleration of the merry-go-round
= The linear velocity of the merry-go-round
t = The time of motion
For the boy, we have;
Where;
= The rotational inertia of the boy
= The angular acceleration of the boy
= The linear velocity of the boy
t = The time of motion
When the boy jumps on the merry-go-round, we have;
Which gives;
From which we have;
The velocity of the merry-go-round, , after the boy hops on the merry-go-round = 1.5 m/s.