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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta P = \rho g h\] | Hydrostatic equilibrium: the pressure difference \(\Delta P\) supports a column of fluid of height \(h\), density \(\rho\), under gravity \(g\). |
| 2 | \[\Delta P = P_{\text{atm}} – P_{\text{suction}}\] | The rise occurs because atmospheric pressure \(P_{\text{atm}}\) exceeds the bulb’s suction pressure \(P_{\text{suction}}\). |
| 3 | \[\rho = \frac{\Delta P}{g h}\] | Algebraically isolate \(\rho\) from the hydrostatic relation. |
| 4 | \[\Delta P = 101{,}300\,\text{Pa} – 99{,}800\,\text{Pa} = 1{,}500\,\text{Pa}\] | Use standard atmospheric pressure (\(101{,}300\,\text{Pa}\)) and the given suction pressure to find \(\Delta P\). |
| 5 | \[\rho = \frac{1{,}500\,\text{Pa}}{(9.8\,\text{m/s}^2)(0.09\,\text{m})}\] | Substitute \(\Delta P = 1{,}500\,\text{Pa}\), \(g = 9.8\,\text{m/s}^2\), and \(h = 9\,\text{cm} = 0.09\,\text{m}\). |
| 6 | \[\rho \approx 1.7 \times 10^{3}\,\text{kg/m}^3\] | Calculate: \(\rho \approx 1{,}700\,\text{kg/m}^3\), which rounds to \(1.7 \times 10^{3}\,\text{kg/m}^3\). |
| 7 | \[\boxed{\rho \approx 1.7 \times 10^{3}\,\text{kg/m}^3}\] | Box the final density of the turkey juice. |
Just ask: "Help me solve this problem."
Which of the following statements is an expression of the equation of continuity?
A helium-filled balloon is attached by a string of negligible mass to a small \(0.015 \ \text{kg}\) object that is just heavy enough to keep the balloon from rising. The total mass of the balloon, including the helium, is \(0.0050 \ \text{kg}\). The density of air is \(\rho_{\text{air}} = 1.29 \ \text{kg/m}^3\), and the density of helium is \(\rho_{\text{He}} = 0.179 \ \text{kg/m}^3\). The buoyant force on the \(0.015 \ \text{kg}\) object is small enough to be negligible.
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 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 block of weight \( W \) is floating in water, and one-third of the block is above the surface of the water. Which of the following correctly describes the magnitude \( F \) of the force that the block exerts on the water and explains why \( F \) has that value?
\(1.7 \times 10^{3}\,\text{kg/m}^{3}\)
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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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