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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."
The difference in pressure between the atmosphere and the human lungs is \( 1.05 \times 10^5 \) \( \text{Pa} \). What is the longest straw you could use to draw up milk whose density is \( 1030 \) \( \text{kg/m}^3 \)?
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 pump, submerged at the bottom of a well that is \( 35 \) \( \text{m} \) deep, is used to pump water uphill to a house that is \( 50 \) \( \text{m} \) above the top of the well, as shown to the right. The density of water is \( 1000 \) \( \text{kg/m}^3 \). All pressures are gauge pressures. Neglect the effects of friction, turbulence, and viscosity.
An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
Marc’ favorite ride at Busch Gardens is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a force of \( 72 \) \( \text{N} \) to a \( 30 \)\( \text{cm}^2 \) cylindrical piston, which holds the \( 20,000 \)\( \text{N} \) ride off the ground. What is the area of the piston that holds the ride?
\( Q_{AB} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( v_{AB} = 4.8 \) \( \text{m/s} \)
\( Q_{BC} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( d_{CD} = 0.454 \) \( \text{m} \) \( = 454 \) \( \text{mm} \)
\( v_{CE} = 6.4 \) \( \text{m/s} \)
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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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