Answer:
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Explanation:
We can answer this question by using Kepler's second law of planetary motion, which states that:
"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"
This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).
In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
This is another time to look at Newton's 2nd law of motion:
Net Force = (mass) x (acceleration)
If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:
Net Force = (mass) x (zero)
or Net Force = (zero) .
"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.
So the answer is '<em>yes</em>', and that's why.
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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} =
