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# 2 short questions

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1: Joe has exactly enough paint to paint the surface of a cube whose side length is 2. It turns out that this is also exactly enough paint to paint the surface of a sphere. If the volume of this sphere is {Ksqrt6}/{sqrt(pi)}, then what is K?

2: 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?

Thanks!

May 17, 2018

#1
+101344
+1

2: 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?

Note that the  volume of water carried by  one 10 in  diameter pipe of  some length, L, is :

pi (10/2)^2  * L  =  25 L  pi    units ^3

Note that the  volume of water carried  by one 2 in diameter pipe of the same length, L, is :

pi (2/2)^2 * L  =  L pi  units^3

So.....we would need 25 times the number of 2 in diameter pipes of some length L to carry the same amount of water as one 10 in diameter pipe of the same length

May 17, 2018
#2
+101344
+1

1: Joe has exactly enough paint to paint the surface of a cube whose side length is 2. It turns out that this is also exactly enough paint to paint the surface of a sphere. If the volume of this sphere is {Ksqrt6}/{sqrt(pi)}, then what is K?

The total surface area  of the cube  is   6 * 2^2  = 24 units^2

So......for the sphere, we can find the radius as

24   = 4 pi ^ r^2

6/pi  = r^2

√[6/pi]  = r

So.....the volume of the sphere  is

K√[6/pi]  =  (4/3) pi * [ √[6/pi] ] ^3

K = (4/3) pi  * √[6/pi]^2

K = (4/3) pi *[  6 / pi ]

K = (4/3) * 6

K  =  8

May 17, 2018
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+1438
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Thank you!

AnonymousConfusedGuy  May 17, 2018