| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_{BC} = \frac{\pi}{4} \times (0.6)^2 \] | Calculate the cross-sectional area of section BC using the diameter \( 600 \text{ mm} = 0.6 \text{ m} \). |
| 2 | \[ A_{BC} = 0.2827 \ \text{m}^2 \] | Evaluate the expression to get the area. |
| 3 | \[ Q_{BC} = A_{BC} \times v_{BC} \] | Use the formula for flow rate, \( Q = A \cdot v \). |
| 4 | \[ Q_{BC} = 0.2827 \times 1.2 \] | Substitute \( v_{BC} = 1.2 \ \text{m/s} \) into the equation. |
| 5 | \[ \boxed{Q_{BC} = 0.3393 \ \text{m}^3/\text{s}} \] | Calculate to find \( Q_{BC} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q_{AB} = Q_{BC} \] | Using the law of mass conservation, \( Q_{AB} = Q_{BC} \) since no other flows are present between A and C. |
| 2 | \[ \boxed{Q_{AB} = 0.3393 \ \text{m}^3/\text{s}} \] | \( Q_{BC} \) was calculated earlier as \( 0.3393 \ \text{m}^3/\text{s} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_{AB} = \frac{\pi}{4} \times (0.3)^2 \] | Calculate the cross-sectional area of section AB using \( 300 \text{ mm} = 0.3 \text{ m} \). |
| 2 | \[ A_{AB} = 0.0707 \ \text{m}^2 \] | Evaluate the expression for area. |
| 3 | \[ v_{AB} = \frac{Q_{AB}}{A_{AB}} \] | Rearrange the formula \( Q = A \cdot v \) to solve for \( v \). |
| 4 | \[ v_{AB} = \frac{0.3393}{0.0707} \] | Substitute \( Q_{AB} \) and \( A_{AB} \) into the equation. |
| 5 | \[ \boxed{v_{AB} = 4.8 \ \text{m/s}} \] | Evaluate to find \( v_{AB} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q_{CD} = \frac{Q_{AB}}{1.5} \] | From \( Q_{AB} = Q_{CD} + Q_{CE} \), solve for \( Q_{CD} \), knowing that \( Q_{CE} = .5Q_{CD} \). |
| 2 | \[ Q_{CD} = \frac{0.3393}{1.5} \] | Substitute \( Q_{AB} = 0.3393 \ \text{m}^3/\text{s} \). |
| 3 | \[ Q_{CD} = 0.2262 \ \text{m}^3/\text{s} \] | Calculate to find \( Q_{CD} \). |
| 4 | \[ A_{CD} = \frac{Q_{CD}}{v_{CD}} \] | Rearrange \( Q = A \cdot v \) to solve for \( A \). |
| 5 | \[ A_{CD} = \frac{0.2262}{1.4} \] | Substitute \( Q_{CD} \) and \( v_{CD} = 1.4 \ \text{m/s} \). |
| 6 | \[ A_{CD} = 0.1616 \ \text{m}^2 \] | Evaluation to find \( A_{CD} \). |
| 7 | \[ d_{CD} = \sqrt{\frac{4 \times A_{CD}}{\pi}} \] | Calculate the diameter from the area. |
| 8 | \[ d_{CD} = \sqrt{\frac{4 \times 0.1616}{\pi}} \] | Substitute \( A_{CD} \) into the equation. |
| 9 | \[ \boxed{d_{CD} = 0.454 \ \text{m}} \] | Calculate the diameter \( d_{CD} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q_{CE} = 0.5Q_{CD} \] | From given condition \( Q_{CE} = 0.5Q_{CD} \). |
| 2 | \[ Q_{CE} = 0.5 \times 0.2262 \] | Use previously calculated \( Q_{CD} \). |
| 3 | \[ Q_{CE} = 0.1131 \ \text{m}^3/\text{s} \] | Evaluate to find \( Q_{CE} \). |
| 4 | \[ A_{CE} = \frac{\pi}{4} \times (0.15)^2 \] | Calculate \( A_{CE} \) with \( 150 \text{ mm} = 0.15 \text{ m} \). |
| 5 | \[ A_{CE} = 0.0177 \ \text{m}^2 \] | Evaluate for \( A_{CE} \). |
| 6 | \[ v_{CE} = \frac{Q_{CE}}{A_{CE}} \] | Rearrange \( Q = A \cdot v \) to solve for \( v \). |
| 7 | \[ v_{CE} = \frac{0.1131}{0.0177} \] | Substitute \( Q_{CE} \) and \( A_{CE} \). |
| 8 | \[ \boxed{v_{CE} = 6.4 \ \text{m/s}} \] | Calculate \( v_{CE} \). |
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An ideal fluid flows through a pipe with radius \( Q \) and flow speed \( V \). If the pipe splits up into three separate paths, each with radius \( \frac{Q}{2} \), what is the flow speed through each of the paths?
A liquid flows at a constant flow rate through a pipe with circular cross-sections of varying diameters. At one point in the pipe, the diameter is \(2\) \(\text{cm}\) and the flow speed is \(18\) \(\text{m/s}\). What is the flow speed at another point in this pipe, where the diameter is \(3\) \(\text{cm}\).
A student designs an experiment to determine the density of an unknown fluid. The student pours the fluid into a graduated cylinder and attaches an object to a force probe. The object has a density greater than the density of the fluid. The student partially submerges the object into the fluid and records both the volume of fluid displaced in the graduated cylinder and the reading on the force probe. The student then submerges the object further and, at each trial, records the new values of displaced volume and force probe reading until the object is fully submerged. The student constructs a graph of force probe reading (vertical axis) as a function of volume of fluid displaced (horizontal axis). Which of the following statements correctly describes how a feature of this graph is related to the density of the fluid?
A solid plastic cube with uniform density (side length = \(0.5\) \(\text{m}\)) of mass \(100\) \(\text{kg}\) is placed in a vat of fluid whose density is \(1200\) \(\text{kg/m}^3\). What fraction of the cube’s volume floats above the surface of the fluid?
The large piston in a hydraulic lift has a radius of \( 250 \) \( \text{cm}^2 \). What force must be applied to the small piston with a radius of \( 25 \) \( \text{cm}^2 \) in order to raise a car of mass \( 1500 \) \( \text{kg} \)?
A spherical balloon has a radius of \(7.15\) \(\text{m}\) and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of \(930\) \(\text{kg}\)?
Take the density of helium and air to be \(0.18\) \(\text{kg/m}^3\) and \(1.24\) \(\text{kg/m}^3\), respectively.
A cylindrical tank of water (height \( H \)) is punctured at a height \( h \) above the bottom. How far from the base of the tank will the water stream land (in terms of \( h \) and \( H \))? What must the value of \( h \) be such that the distance at which the stream lands will be equal to \( H \)?
A Venturi meter is a device used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed \( v_2 \) through a horizontal section of pipe with a cross-sectional area \( A_2 = 542 \) \( \text{cm}^2 \). The gas has a density of \( 1.35 \) \( \text{kg/m}^3 \). The Venturi meter has a cross-sectional area of \( A_1 = 215 \) \( \text{cm}^2 \) and has been substituted for a section of the larger pipe. The pressure difference between the two sections \( P_2 – P_1 = 145 \) \( \text{Pa} \).
In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
Johnny the auto mechanic is raising a \( 1200 \) \( \text{kg} \) car on her hydraulic lift so that she can work underneath. If the area of the input piston is \( 12 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
\( Q_{AB} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( v_{AB} = 4.8 \) \( \text{m/s} \)
\( Q_{BC} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( d_{CD} = 0.454 \) \( \text{m} \) \( = 454 \) \( \text{mm} \)
\( v_{CE} = 6.4 \) \( \text{m/s} \)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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