Water flows into a house at a velocity of 15 m/s through a pipe that has a radius of 0.40 m. The water then flows through a smaller pipe at a velocity of 45 m/s. What is the radius of the smaller pipe?
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Well, well, we've got a water flow conundrum here! Let's tackle it together, shall we?
We can use the principle of conservation of mass to solve this problem. According to this principle, the mass of water entering the house should be equal to the mass of water exiting the smaller pipe. Since the density of water remains constant, we can equate the two pipe's flow rates.
The flow rate (Q) of water is given by the formula Q = A * v, where A denotes the cross-sectional area of the pipe and v represents the velocity of the water.
For the first pipe, with a velocity (v1) of 15 m/s and radius (r1) of 0.40 m, the flow rate (Q1) is:
Q1 = π * r1^2 * v1
Now, for the smaller pipe, let's find the radius (r2) using the given velocity (v2) of 45 m/s. The flow rate (Q2) for the smaller pipe is:
Q2 = π * r2^2 * v2
Since the flow rates are equal, we can write: Q1 = Q2
Here's the equation: π * r1^2 * v1 = π * r2^2 * v2
Now, if we simplify this equation, we can solve for r2:
(r2^2) = (r1^2 * v1 * v2) / (v2) = (r1^2 * v1) / (v2)
Finally, take the square root of both sides to find the radius of the smaller pipe (r2):
r2 = √((r1^2 * v1) / (v2))
Plug in the given values for r1, v1, and v2, and you'll find the radius of the smaller pipe!
To find the radius of the smaller pipe, we can use the principle of continuity, which states that the product of the cross-sectional area and the velocity of a fluid at one point in a pipe is equal to the product of the cross-sectional area and the velocity of the fluid at another point in the same pipe.
The equation for the continuity principle is:
A1v1 = A2v2
Where A1 and v1 are the cross-sectional area and velocity at the first point, and A2 and v2 are the cross-sectional area and velocity at the second point.
In this case, we know the velocity at the first point (15 m/s) and the radius at the first point (0.40 m). We want to find the radius at the second point.
First, let's find the cross-sectional area at each point using the formula for the area of a circle:
A = πr^2
For the first point:
A1 = π(0.40 m)^2
For the second point, we'll call the radius of the smaller pipe r2:
A2 = πr2^2
Now, we can rewrite the continuity equation using these values:
π(0.40 m)^2(15 m/s) = πr2^2(45 m/s)
Simplifying:
(0.40 m)^2(15 m/s) = r2^2(45 m/s)
(0.16 m^2)(15 m/s) = r2^2(45 m/s)
(2.4 m^2/s) = r2^2(45 m/s)
Now, we can solve for r2 by dividing both sides by (45 m/s):
(2.4 m^2/s) / (45 m/s) = r2^2
0.05333 m = r2^2
Finally, we take the square root of both sides to find the radius:
r2 = √(0.05333 m)
r2 ≈ 0.231 m
Therefore, the radius of the smaller pipe is approximately 0.231 meters.