+1206 A pipe with inside diameter 10'' is to carry water from a reservoir to a small town in an arid land. Neglecting the friction and turbulence of the water against the inside of the pipes, what is the minimum number of 2''-inside-diameter pipes of the same length needed to carry the same volume of water to the arid town?
+36915 The sum of the cross sectional area of the 2" pipes must equal the cross sectional area of the larger 10" pipe
Cross sectional area = pi r^2
x pi r^2 = pi 5^2 where x = # of 2" pipes pi 5^2 = area of 10" pipe
x pi (1)^2 = pi 5^2 divide both sides by pi
x = 5^2
x = 25 Yep, you'll need 25 of the 2" pipes to match the 10" pipe !
+36915 The sum of the cross sectional area of the 2" pipes must equal the cross sectional area of the larger 10" pipe
Cross sectional area = pi r^2
x pi r^2 = pi 5^2 where x = # of 2" pipes pi 5^2 = area of 10" pipe
x pi (1)^2 = pi 5^2 divide both sides by pi
x = 5^2
x = 25 Yep, you'll need 25 of the 2" pipes to match the 10" pipe !
