A liquid (ñ = 1.65 g/cm3) flows through a horizontal pipe of varying cross section as in the figure below. In the first section, the cross-sectional area is 10.0 cm2, the flow speed is 288 cm/s, and the pressure is 1.20 105 Pa. In the second section, the cross-sectional area is 3.50 cm2.

(a) Calculate the smaller section's flow speed.
m/s

(b) Calculate the smaller section's pressure.

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To answer this question, we can apply the principle of conservation of mass and Bernoulli's equation.

(a) According to the principle of conservation of mass, the mass flow rate remains constant throughout the pipe. The mass flow rate (m_dot) is given by the equation:

m_dot = ρ * A * v

where ρ is the density of the liquid, A is the cross-sectional area of the pipe, and v is the flow speed.

In the first section of the pipe:
A1 = 10.0 cm^2 = 0.001 m^2
v1 = 288 cm/s = 2.88 m/s
ρ = 1.65 g/cm^3 = 1650 kg/m^3

Using the equation for mass flow rate, we can calculate the mass flow rate in the first section (m_dot1):

m_dot1 = ρ * A1 * v1

In the second section of the pipe:
A2 = 3.50 cm^2 = 0.00035 m^2
v2 is the flow speed that we need to find.

Using the principle of conservation of mass, we can equate the mass flow rate in the two sections:

m_dot1 = m_dot2

ρ * A1 * v1 = ρ * A2 * v2

Plugging in the known values:

1650 kg/m^3 * 0.001 m^2 * 2.88 m/s = 1650 kg/m^3 * 0.00035 m^2 * v2

Solving for v2:

v2 = (1650 kg/m^3 * 0.001 m^2 * 2.88 m/s) / (1650 kg/m^3 * 0.00035 m^2)

v2 ≈ 7.94 m/s

Therefore, the flow speed in the smaller section is approximately 7.94 m/s.

(b) To calculate the pressure in the smaller section, we can use Bernoulli's equation:

P1 + 0.5 * ρ * v1^2 + ρ * g * h1 = P2 + 0.5 * ρ * v2^2 + ρ * g * h2

where P1 and P2 are the pressures in the larger and smaller sections respectively, v1 and v2 are the flow speeds, ρ is the density of the liquid, g is the acceleration due to gravity, and h1 and h2 are the heights of the two sections.

Since both sections are at the same height and the problem does not provide any information about a height difference, we can drop the terms ρ * g * h1 and ρ * g * h2 from the equation.

P1 + 0.5 * ρ * v1^2 = P2 + 0.5 * ρ * v2^2

Rearranging the equation, we can solve for P2:

P2 = P1 + 0.5 * ρ * (v1^2 - v2^2)

Plugging in the known values:

P2 = 1.20 * 10^5 Pa + 0.5 * 1650 kg/m^3 * (2.88 m/s)^2 - (7.94 m/s)^2

Calculating the pressure:

P2 ≈ 1.20 * 10^5 Pa - 10984.25 Pa

P2 ≈ 1.19 * 10^5 Pa

Therefore, the pressure in the smaller section is approximately 1.19 * 10^5 Pa.