(C)
Explanation:

If g is only 1/6 on another planet, then


Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
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B1 at 20km/h
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V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.
The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.
The given parameters;
- time of fall of the first ball, t = 1 s
- time of fall of the second ball, t = 3 s
The distance traveled by each ball is calculated using the second equation of motion as shown below.
The distance traveled by the first ball is calculated as follows;

The distance traveled by the second ball is calculated as follows;

Thus, the second ball traveled a greater distance because it spent more time in motion.
Learn more here:brainly.com/question/5868480
Answer:
<u>Magnitude</u>
Explanation:
Each value in nature has a number part, called its magnitude and a dimension called its unit.
For example,
The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.