The answer to the statement is true because the day is of the logical proportion it has to be time.
Answer:
Part a) 
Part b) 
Explanation:
Part a) what is its frequency, in rev/s
we have that
An old-fashioned LP record rotates at 33 1/3 RPM
so

Convert mixed number to an improper fraction

Remember that

Convert rev/min to rev/sec

Simplify

Part b) what is it period, in seconds
we know that
The period is the reciprocal of the frequency
therefore
the frequency is

Answer:
a) 
b) The second runner will win
c) d = 10.54m
Explanation:
For part (a):

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:


Since it takes less time to the second runner to cross the finish line, we can say the she won the race.
For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:
X1 = V1*t2 = 239.46m Since the finish line was 250m away:
d = 250m - 239.46m = 10.54m
Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.