Water (density = 1 × 10³ kg/m³)flows at 11 m/s through a pipe with radius 0.025 m.The pipe goes up to the second floor of the building,2.5 m higher,and the pressure remains unchanged.What is the radius of the pipe on the second floor? A) 0.01 m B) 0.028 m C) 0.001 m D) 0.18 m E) 0.023 m

Question

Quizplus · Accepted Answer

Since the pressure remains unchanged, we can use Bernoulli's equation:

1/2pv1^2 + pgh1 = 1/2pv2^2 + pgh2

Where p is density, v is velocity, g is gravity, and h is height.

Since we have the same pressure at both heights, we can cancel out the pressure terms. Also, since the pipe has a constant diameter, the fluid will be incompressible and have constant density at both heights.

Therefore, we have:

1/2v1^2 + gh1 = 1/2v2^2 + gh2

Solving for v2 (velocity on the second floor), we get:

v2 = sqrt(v1^2 - 2gh)

Plugging in the given values, we get:

v2 = sqrt((11 m/s)^2 - 2(9.81 m/s^2)(2.5 m)) = 10.18 m/s

Now we can use the continuity equation:

A1v1 = A2v2

Where A is cross-sectional area.

Solving for A2 (cross-sectional area on the second floor), we get:

A2 = A1(v1/v2) = pi(0.025 m)^2(11 m/s/10.18 m/s) = 0.028 m^2

Finally, we can solve for the radius on the second floor:

A2 = pi(r2^2) => r2 = sqrt(A2/pi) = 0.028 m.

Therefore, the correct answer is B.

Water (Density = 1 × 103 Kg/m3)flows at 11 M/s

Water (Density = 1 × 10³ Kg/m³)flows at 11 M/s