Answer:C (198 seconds)
Explanation: The cyclist makes the first lap in (180.00 - 6.00) = 174.00 seconds. The average time per lap for all three circuits is (600.00 - 6.00) = 594/3 = 198 seconds.
Answer:
Required diameter of hose pipe = 0.2864 mm
Solution:
From the continuity eqn, the fluid flow rate is given by:
Av = 
where
A = cross-sectional area = 
r = hose pipe radius
v = velocity of gas
Also, 
Using:
1 gallon = 3.854 l
1 mile = 1609.34 m

Therefore,






The diameter of the hose pipe = 2r = 
The diameter of the hose pipe = 
To develop this problem we will apply the concepts related to the Electromagnetic Force. The magnetic force can be defined as the product between the free space constant, the current (of each cable) and the length of these, on the perimeter of the cross section, in this case circular. Mathematically it can be expressed as,

Here,
= Permeability free space
I = Current
L = Length
d= Distance between them
Our values are,




Rearranging the previous equation to find the current,





Therefore the current in the rods is 210.6A
Geology because It is the study of processes that shape Earth.