Question 24
1 answer:
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Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Correct order, from lowest potential energy to highest potential energy:
E - C - D - B - A
Explanation:
The gravitational potential energy of the car is given by:
where
m is the car's mass
g is the gravitational acceleration
h is the height of the car relative to the ground
In the formula, we see that m and g are constant, so the potential energy of the car depends only on its height above the ground, h. The higher the car from the ground, the larger its potential energy. Therefore, the position with least potential energy will be E, since the height is the minimum. Then, C will have more potential energy, because the car is at higher position, and so on: the position with greatest potential energy is A, because the height of the car is maximum.