SOLUTION: The amount of silt carried by a stream is directly proportional to the sixth power of its velocity. A certain stream carries 150 tons of silt per day. Determine the amount of silt

Question 884738: The amount of silt carried by a stream is directly proportional to the sixth power of its velocity. A certain stream carries 150 tons of silt per day. Determine the amount of silt it would carry if its velocity were doubled.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the original formula is:
S1 = k * v^6

when you double the velocity, the revised formula becomes:
S2 = k * (2v)^6 which becomes:
S2 = k * 2^6 * v^6 which becomes:
S2 = k * 64 * v^6 which can also be written as:
S2 = 64 * k * v^6

the ratio of the revised formula to the original formula is:

S2/S1 = (64 * k * v^6) / (k * v^6)
the (k * v^6) in the numerator and the denominator cancel out and you are left with:
S2/S1 = 64

this means that S2 is 64 times as large as S1
this means that. if S1 = 150, then S2 = 150 * 64 = 9600

the answer to the question is that the amount of silt carried per day would be equal to 9600 tons.

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