A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is 96.0 kg, and the hill is
inclined at 17.0° with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?
1 answer:
Answer:
The force applied 275 N in a direction parallel to the hill
Explanation:
Newton's second law is adequate to work this problem, in the annex we can see a free body diagram, where the weight (W) is vertical, the friction force (fr) is parallel to the surface and the normal (N ) is perpendicular to it. In general for these problems a reference system is taken that is parallel to the surface and the Y axis is perpendicular to it.
Let us decompose the weight into its two components, the angle T is taken from the axis and
Wx = W sin θ
Wy = W cos T
We write Newton's second law
∑ F = m a
X axis
The cyclist falls at a constant speed, which implies that the acceleration is zero
fr - W sin θ = 0
fr = mg sin θ
fr = 96 9.8 without 17
fr = 275 N
When the cyclist returns to climb the hill, he must apply the same force he has to overcome the friction force that always opposes the movement . The force applied 275 N in a direction parallel to the hill
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Answer:
23376 days
Explanation:
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where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be and its mean distance of from the sun be
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Also let the orbital period of the planet be and its mean distance from the sun be
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Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
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cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
1) Velocity: 9.9 m/s and 14 m/s
The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:
where
v is the final velocity
u = 0 is the initial velocity (the diver starts from rest)
is the acceleration of gravity
s is the displacement
For the diver jumping from 5 m, s = 5 m, so
For the diver jumping from 10 m, s = 10 m, so
2) Time: 1.01 s and 1.43 s
The time of flight of each diver can be found by using the other suvat equation
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For the diver jumping from 5 m, s = 5 m, so we find
For the diver jumping from 10 m, s = 10 m, so we find