Chapter 11: Problem 58
Water flows straight down from an open faucet. The cross- sectional area of the faucet is \(1.8 \times 10^{-4} \mathrm{m}^{2},\) and the speed of the water is 0.85 \(\mathrm{m} / \mathrm{s}\) as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point 0.10 \(\mathrm{m}\) below the faucet.
Short Answer
Step by step solution
Understand the Problem
Apply Bernoulli's Equation
Use Continuity Equation
Calculate Velocity Change Due to Gravity
Substitute Values
Solve for New Cross-Sectional Area
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bernoulli's Equation
The equation can be expressed as:\[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]Where:
- \( P \) is the fluid's pressure
- \( \rho \) is the fluid density
- \( v \) is the fluid velocity
- \( g \) is acceleration due to gravity
- \( h \) is the height of the fluid
Continuity Equation
The mathematical form of the continuity equation is:\[ A_1 v_1 = A_2 v_2 \]Where:
- \( A_1 \) and \( v_1 \) are the initial area and velocity at the faucet
- \( A_2 \) and \( v_2 \) are the area and velocity at the point of interest
Gravitational Potential Energy
We calculate GPE using the formula:\[ \text{GPE} = mgh \]Where:
- \( m \) is mass
- \( g \) is acceleration due to gravity
- \( h \) is the height
Fluid Velocity
To find the new velocity at a certain depth, we use the kinematic equation:\[ v_2^2 = v_1^2 + 2gh \]Where:
- \( v_1 \) is initial velocity
- \( v_2 \) is the final velocity
- \( g \) is the acceleration due to gravity
- \( h \) is the change in height