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Question 80080: 1. Hydrodynamics is the branch of physics that studies the behavior of liquids that are in motion. For example, The Principal of Continuity in Liquid Flow states that the velocity of a liquid flowing through a pipe increases as the cross-sectional area of the pipe decreases, and decreases as the cross sectional area of the pipe increases.
On a given day, a fire department uses the same pump to put out two fires. The rate of water is given by the rational function:
r = 1250/A
where r is the rate of water in gallons per minute, and A is the cross sectional area of the hose in square inches. Does the function obey the Principle of Continuity in Liquid Flow?
2. During the morning fire, the fire department used a hose with a cross sectional area of 5 square inches. What was the velocity of the water?
3. During the evening fire, the velocity of the water was 100 gallons per minute. What was the cross sectional area of the hose?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Given:
.
r = 1250/A
.
(1) Does this follow the Principle of Continuity which states that the velocity (r) of a liquid
flowing through a pipe increases as the cross-sectional area of the pipe decreases,
and decreases as the cross sectional area of the pipe increases.
.
The answer is that it does. Try it yourself. Suppose the area is 1 square inch. What is
the velocity (r)? Substitute 1 for A and you find that r = 1250/1 = 1250 gal per min.
.
Next increase the Area to 2 square inches. Substitute 2 for A and you find that r, the
velocity is now: r = 1250/2 = 625 gal per min. So as the Area got bigger (increased),
the velocity got smaller (decreased) ... exactly what the Principle of Continuity
said it should do.
.
(2) Given that the hose for the morning fire had a cross-sectional area of 5 square
inches, what was the velocity of the water? Just substitute 5 for A and you get:
.
r = 1250/5 = 250 gallons per minute
.
(3) During the evening fire, the velocity of the water was 100 gallons per minute.
Substitute this value for r in the equation and solve for A, the cross-sectional area:
.
100 = 1250/A
.
Multiply both sides of the equation by A to eliminate the denominator of A that appears
on the right side:
.
100A = 1250
.
Next divide both sides of the equation by 100 to find A:
.
A = 1250/100 = 12.5 square inches
.
Hope this helps you to understand the problem a little better.
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