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| Derivation or Formula | Reasoning |
|---|---|
| \[W_{\text{air}} = \rho_{\text{rock}} V g\] | This is the full weight of the rock in air, where \(\rho_{\text{rock}}\) is the rock density, \(V\) is its volume, and \(g\) is gravitational acceleration. |
| \[W_{\text{water}} = \rho_{\text{rock}}Vg – \rho_{\text{water}}Vg\] | The apparent weight when submerged is the true weight minus the buoyant force, which equals the weight of the displaced water \((\rho_{\text{water}}Vg)\). |
| \[\rho_{\text{rock}}Vg – \rho_{\text{water}}Vg = \tfrac{1}{2}\,\rho_{\text{rock}}Vg\] | The problem states the rock weighs twice as much in air as in water, so \(W_{\text{water}} = \tfrac{1}{2}W_{\text{air}}\). Substitute \(W_{\text{air}}\) above. |
| \[\rho_{\text{rock}} – \rho_{\text{water}} = \tfrac{1}{2}\,\rho_{\text{rock}}\] | Canceling the common factors \(Vg\) from both sides simplifies the equation. |
| \[2\,\rho_{\text{rock}} – 2\,\rho_{\text{water}} = \rho_{\text{rock}}\] | Multiply both sides by 2 to clear the fraction. |
| \[\rho_{\text{rock}} = 2\,\rho_{\text{water}}\] | Solve for \(\rho_{\text{rock}}\) by subtracting \(\rho_{\text{rock}}\) from both sides. |
| \[\rho_{\text{rock}} = 2 \times 1000\,\text{kg/m}^3 = \boxed{2000\,\text{kg/m}^3}\] | Assuming the density of water is \(1000\,\text{kg/m}^3\), the rock density becomes \(2000\,\text{kg/m}^3\) (to 3 significant figures). |
| Derivation or Formula | Reasoning |
|---|---|
| \[\rho_{\text{avg}} = p\,\rho_{\text{quartz}} + (1-p)\times0 = p\,\rho_{\text{quartz}}\] | The rock is part solid quartz and part hollow (air). Since the density of air is negligible, \(\rho_{\text{avg}}\) is just the fraction \(p\) of quartz times the density of quartz \(\rho_{\text{quartz}}\). |
| \[p = \tfrac{\rho_{\text{avg}}}{\rho_{\text{quartz}}}\] | Solve for \(p\) (the fraction that is solid quartz) using the average density, which we determined as \(2000\,\text{kg/m}^3\), and the given quartz density \(2660\,\text{kg/m}^3\). |
| \[p = \tfrac{2000}{2660} \approx 0.752\] | Calculate \(p\) to determine the fraction of the rock that is quartz. |
| \[\text{Percentage hollow} = (1-p) \times 100\% \approx (1-0.752) \times 100\% \approx \boxed{24.8\%}\] | The remainder of the rock (\(1-p\)) is hollow. Converting to a percentage gives about 24.8\% hollow, to 3 significant figures. |
Just ask: "Help me solve this problem."
An air mattress pump blows air above a beach ball at \( 8 \) \( \text{m/s} \). The air below the beach ball is moving at \( \approx 0 \) \( \text{m/s} \). Assuming the beach ball diameter is \( 0.1 \) \( \text{m} \), meaning the areas for the top \& bottom are each \( \approx 0.03 \) \( \text{m}^2 \), and the density of air is \( 1 \) \( \text{kg/m}^3 \), what is the lift force on the beach ball?
The radius of the aorta is about \( 1 \) \( \text{cm} \) and the blood flowing through it has a speed of about \( 30 \) \( \frac{\text{cm}}{\text{s}} \). Calculate the average speed of the blood in the capillaries given the total cross section of all the capillaries is about \( 2000 \) \( \text{cm}^2 \).
The diagram above shows a hydraulic chamber with a spring \( (k_s = 1250 \, \text{N/m}) \) attached to the input piston and a rock of mass \( 55.2 \, \text{kg} \) resting on the output plunger. The input piston and output plunger are at about the same height, and each has negligible mass. The chamber is filled with water.
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 \( 1030 \) \( \text{kg/m}^3 \)?
The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is \( 2.5 \) \( \text{m} \), and the puncture is \( 1 \) \( \text{m} \) above the ground level, what is the efflux speed of the water?
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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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