In a machine, an ideal fluid with a density of 0.9*10^3 kg/m^3 and a pressure of 1.3*10^5 Pa flows at a velocity of 6.0 m/s through a level tube with a readius of 0.5 cm. This tube connects to a level tube that has a radius of 1.5 cm. How fast does the water flow in the larger tube?
a. 54 m/s
b. 18 m/s
c. 0.67 m/s
d. 0.33 m/s
An ideal fluid is one that is incompressible and has no viscosity. Ideal fluids do not actually exist, but sometimes it is useful to consider what would happen to an ideal fluid in a particular fluid flow problem in order to simplify the problem.
If it incompressible, then consider the law of mass flow.
Density*velocity*area=constant
or 6m/s*(1/2)^2 *PI=V*(3/2)^2 * PI
solve for Velocity V.
To solve for the velocity in the larger tube, we can use the law of mass flow, which states that the product of the density, velocity, and cross-sectional area of a fluid remains constant.
First, let's calculate the cross-sectional areas of the two tubes. The radius of the smaller tube is 0.5 cm, which is equal to 0.005 m. The radius of the larger tube is 1.5 cm, which is equal to 0.015 m.
The cross-sectional area of the smaller tube can be found using the formula A = πr^2, where r is the radius:
A1 = π(0.005)^2 = 0.0000785 m^2
Similarly, the cross-sectional area of the larger tube is:
A2 = π(0.015)^2 = 0.0007069 m^2
According to the law of mass flow, the product of the density, velocity, and cross-sectional area is constant. We can set up an equation with this information:
density1 * velocity1 * A1 = density2 * velocity2 * A2
Plugging in the given values:
(0.9*10^3 kg/m^3) * (6.0 m/s) * (0.0000785 m^2) = density2 * velocity2 * (0.0007069 m^2)
Simplifying the equation, we can solve for velocity2:
velocity2 = ((0.9*10^3 kg/m^3) * (6.0 m/s) * (0.0000785 m^2)) / ((0.0007069 m^2))
velocity2 ≈ 0.0997 m/s
Therefore, the water flows at a velocity of approximately 0.0997 m/s in the larger tube. Since none of the given answer choices match this result exactly, the closest option is 0.33 m/s (option d).