A liquid (ρ = 1.65 g/cm^3) flows through two horizontal sections of tubing joined end to end. In the first section the cross-sectional area is 10.0 cm^2, the flow speed is 245 cm/s, and the pressure is 1.20 105 Pa. In the second section the cross-sectional area is 2.50 cm^2.
To answer this question, we can make use of the principle of continuity, which states that the flow rate of an incompressible liquid remains constant as it moves through different sections of tubing.
The flow rate (Q) is given by the equation:
Q = A1 * v1 = A2 * v2
Where:
Q = flow rate (constant)
A1 = cross-sectional area of the first section
v1 = flow speed in the first section
A2 = cross-sectional area of the second section
v2 = flow speed in the second section
Given:
A1 = 10.0 cm^2
v1 = 245 cm/s
A2 = 2.50 cm^2
To find v2, we can rearrange the equation:
v2 = (A1 * v1) / A2
Now, let's plug in the given values:
v2 = (10.0 cm^2 * 245 cm/s) / 2.50 cm^2
To simplify the units, we'll convert cm^2 to m^2 and cm/s to m/s:
v2 = (10.0 * 10^-4 m^2 * 245 m/s) / (2.50 * 10^-4 m^2)
Simplifying further:
v2 = 9.8 m/s
So, the flow speed in the second section of tubing is 9.8 m/s.
Next, we can calculate the pressure in the second section of tubing using Bernoulli's equation. Bernoulli's equation states that the total energy of a fluid remains constant along a streamline. It can be expressed as:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2
Where:
P1 = pressure in the first section
ρ = density of the liquid
v1 = flow speed in the first section
P2 = pressure in the second section
v2 = flow speed in the second section
Given:
P1 = 1.20 * 10^5 Pa
ρ = 1.65 g/cm^3 = 1650 kg/m^3 (converting from g/cm^3 to kg/m^3)
v1 = 245 cm/s = 2.45 m/s (converting from cm/s to m/s)
v2 = 9.8 m/s (from previous calculation)
Plugging in the values:
1.20 * 10^5 Pa + (1/2) * 1650 kg/m^3 * (2.45 m/s)^2 = P2 + (1/2) * 1650 kg/m^3 * (9.8 m/s)^2
Simplifying further:
1.20 * 10^5 Pa + 6.00 * 10^2 Pa = P2 + 7.65 * 10^4 Pa
Rearranging the equation to solve for P2:
P2 = 1.20 * 10^5 Pa + 6.00 * 10^2 Pa - 7.65 * 10^4 Pa
P2 = 8.16 * 10^4 Pa
So, the pressure in the second section of tubing is 8.16 * 10^4 Pa.