A liquid (density = 1.65 g/cm^3) flows through two horizontal sections of tubing. In the first section the cross-sectional area is 10.0 cm^2, the flow speed is 275 cm/s, and the pressure is 1.20 x 10^5 Pa. This tubing then rises 50 cm to the second horizontal section which has a cross-sectional area of 2.50 cm^2. Calculate the second section's flow speed and pressure.

You will need to convert all units to "standard metric."

Use the law of continuity of incompressible fluids to get the velocity in the constricted section.

That law says that the product of cross sectional area and velocity must be constant in steady state. The law tells you that the flow velocity in the 2.50 cm^2 section is ten times higher than it is in the 10 cm^2 section.

Knowing the velocity in both places, use Bernoulli's equation to get the pressure change.

p + (1/2)*rho*V^2 + rho*g*H = constant.

H is the height of the tubing and rno is the density. With the above equation, the CHANGE in pressure can be related to the change in V and H.

I know the anwers are 11.0 m/s and 1.83 x 10^4 Pa

I just need to know how to work it out and how to start it

Ok thx. Sounds good. I already found the velocity so just need to get the pressure. Will do the math. And I need help on the other question you helped me on yesterday =P

Well, I'm glad you asked this! Converting units to "standard metric" is a pretty straightforward task, but before we do that, I must insist on telling you a joke.

Why don't scientists trust atoms?

Because they make up everything!

Now, let's get back to business and solve this problem. We can start by converting the given units to the "standard metric" system.

First, let's convert the flow speed from cm/s to m/s. Since 1 meter is equal to 100 centimeters, the flow speed is 2.75 m/s (275 cm/s divided by 100).

Next, let's convert the density from g/cm^3 to kg/m^3. Since 1 kilogram is equal to 1000 grams and 1 cubic meter is equal to 1,000,000 cubic centimeters, the density is 1650 kg/m^3 (1.65 g/cm^3 multiplied by 1000).

Now that we have converted the units, we can move on to solving the problem.

Using the principle of conservation of mass, we know that the flow rate in the first section of tubing will be the same as the flow rate in the second section of tubing.

The flow rate (Q) is given by the equation Q = A * v, where A is the cross-sectional area and v is the flow speed.

So, in the first section, the flow rate is (10.0 cm^2) * (2.75 m/s) = 27.5 cm^3/s, which is equivalent to 0.0000275 m^3/s.

In the second section, the flow rate will be the same, so we can use the equation Q = A * v again.

We know that the cross-sectional area in the second section is 2.50 cm^2, which is equivalent to 0.000250 m^2.

Let's call the flow speed in the second section v2. So, the equation becomes:

(0.000250 m^2) * v2 = 0.0000275 m^3/s

Solving for v2, we find that v2 = 0.0000275 m^3/s / 0.000250 m^2 = 0.11 m/s.

Now, let's move on to calculating the pressure in the second section.

We can use Bernoulli's equation, which states that the pressure at any point in a fluid is equal to the pressure at a reference point (in this case, the first section) plus a term related to the fluid's velocity and density.

Mathematically, this equation is written as P + (1/2) * ρ * v^2 = P1 + (1/2) * ρ * v1^2, where P is the pressure, ρ is the density, and v is the velocity.

By rearranging the equation, we can find the pressure in the second section (P2):

P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)

Now, let's plug in the given values:

P1 = 1.20 x 10^5 Pa,
ρ = 1650 kg/m^3 (which we calculated earlier),
v1 = 2.75 m/s,
v2 = 0.11 m/s.

Calculating P2 using the equation, we find:

P2 = 1.20 x 10^5 Pa + (1/2) * (1650 kg/m^3) * ((2.75 m/s)^2 - (0.11 m/s)^2)

After crunching the numbers, P2 turns out to be approximately 1.13 x 10^5 Pa.

So, we have successfully calculated the flow speed and pressure in the second section. I hope you found this explanation helpful, and if you have any more questions, feel free to ask!

To calculate the flow speed and pressure in the second section, we need to apply the principle of continuity and Bernoulli's equation. Here's how we can solve it step by step:

1. Start by converting the given densities to standard metric units. The density of the liquid is 1.65 g/cm^3. To convert it to kg/m^3, we multiply by 1000 since there are 1000 cm in a meter. Therefore, the density is 1650 kg/m^3.

2. Next, convert the given areas to square meters. The first section has a cross-sectional area of 10.0 cm^2. To convert it to m^2, divide by 10000 since there are 10000 cm^2 in a m^2. So, the first section's area is 0.001 m^2. The second section has an area of 2.50 cm^2, which is equivalent to 0.00025 m^2.

3. Now, use the principle of continuity to relate the flow speeds at the two sections. According to the principle of continuity, the product of the cross-sectional area and the flow speed is constant for an incompressible fluid. Thus, A1v1 = A2v2, where A1 and v1 are the area and speed in the first section, and A2 and v2 are the area and speed in the second section.

4. Using the given values, we have (0.001 m^2)(275 cm/s) = (0.00025 m^2)(v2). Now, solve for v2.

0.001 m^2 * 275 cm/s = 0.00025 m^2 * v2
v2 = (0.001 m^2 * 275 cm/s) / (0.00025 m^2)
v2 ≈ 1100 cm/s

Therefore, the flow speed in the second section is approximately 1100 cm/s.

5. Finally, to calculate the pressure in the second section, we can use Bernoulli's equation, which states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline. Since the second section is at a different height from the first, we need to account for the change in potential energy.

Bernoulli's equation can be written as P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2, where P1 and P2 are the pressures, v1 and v2 are the velocities, ρ is the density, g is the acceleration due to gravity, h1 is the height in the first section, and h2 is the height in the second section.

In this case, the liquid is flowing horizontally, so the heights h1 and h2 can be ignored. Bernoulli's equation simplifies to P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2.

We are given P1 = 1.20 x 10^5 Pa and v1 = 275 cm/s (which we can convert to m/s by dividing by 100). Plugging in these values, we have:

1.20 x 10^5 Pa + (1/2)(1650 kg/m^3)(2.75 m/s)^2 = P2 + (1/2)(1650 kg/m^3)(1100 cm/s)^2

Simplifying the equation gives:

1.20 x 10^5 Pa + 3.31 x 10^5 Pa = P2 + 1.40 x 10^8 Pa

Subtracting 1.20 x 10^5 Pa and 3.31 x 10^5 Pa from both sides gives:

P2 = 1.40 x 10^8 Pa - 4.51 x 10^5 Pa
P2 ≈ 1.40 x 10^8 Pa - 4.51 x 10^5 Pa
P2 ≈ 1.40 x 10^8 Pa - 451000 Pa
P2 ≈ 1.40 x 10^8 Pa - 1.40 x 10^5 Pa
P2 ≈ 1.40 x 10^8 Pa - 0.14 x 10^8 Pa
P2 ≈ 1.26 x 10^8 Pa

Therefore, the pressure in the second section is approximately 1.26 x 10^8 Pa.