Note: <em>The question states the time to go upstream is a number of times (not explicitly written) the time to go downstream. We'll assume a general number N</em>
Answer:

Explanation:
<u>Relative Speed</u>
If a boat is going upstream against the water current, the true speed of motion is
, being
the speed of the boat and
the speed of the water. If the boat is going downstream, the true speed becomes
.
The question states the time to go upstream is a number of times N (not explicitly written) the time to go downstream. The speed of an object is computed as

Where x is the distance traveled and t the time taken for that. The time can be computed by

If
is the time for the upstream travel and
is the time for the downstream travel, then

Siince the same distance x= 10 miles is traveled in both cases:

Simplifying and rearrangling

Operating

Solving for 



If N=3

We can use the required value of N to compute the speed of the boat as explained
Answer:
c. Position C
Explanation:
As the magnetic force is the strongest at the end of the poles and it decreases when it comes to its centre.
You need to add more information
Answer:
Gases: Oxygen and Carbon dioxide
Liquids: Water
Solid: Most Metals
Answer:
The forces offset to produce a net force of 8 N
since the ball is 1 kg
F = ma
a = f/m = 8 N/1 kg = 8 (kg m)/(s^2 kg) = 8 m/s^2