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| 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 \( v_{CE} \). |
Just ask: "Help me solve this problem."
A fluid flows through the two sections of cylindrical pipe shown in the figure. The narrow section of the pipe has radius \( R \) and the wide section has radius \( 2R \). What is the ratio of the fluid’s speed in the wide section of pipe to its speed in the narrow section of pipe, \( \frac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
A fountain with an opening of radius \( 0.015 \) \( \text{m} \) shoots a stream of water vertically from ground level at \( 6.0 \) \( \text{m/s} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
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?
A solid titanium sphere of radius \( 0.35 \) \( \text{m} \) has a density \( 4500 \) \( \text{kg/m}^3 \). It is held suspended completely underwater by a cable. What is the tension in the cable?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
\( Q_{AB} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( v_{AB} = 4.8 \) \( \text{m/s} \)
\( Q_{BC} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( d_{CD} = 0.454 \) \( \text{m} \) \( = 454 \) \( \text{mm} \)
\( v_{CE} = 6.4 \) \( \text{m/s} \)
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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] |
General Metric Conversion Chart
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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