| Derivation or Formula | Reasoning |
|---|---|
| \[A_1 v_1 = A_2 v_2\] | Continuity for steady flow: the volume flow rate is constant, so \(A_1 v_1 = A_2 v_2\). |
| \[v_1 = \left(\frac{A_2}{A_1}\right)v_2\] | Solve the continuity equation for \(v_1\) in terms of \(v_2\). |
| \[P_2 + \frac{1}{2}\rho v_2^2 = P_1 + \frac{1}{2}\rho v_1^2\] | Bernoulli’s equation for a horizontal pipe: same height so the \( \rho g y \) terms cancel, leaving pressure and kinetic terms. |
| \[P_2 – P_1 = \frac{1}{2}\rho\left(v_1^2 – v_2^2\right)\] | Rearrange Bernoulli to isolate the given pressure difference \(P_2 – P_1\). |
| \[P_2 – P_1 = \frac{1}{2}\rho\left(\left(\frac{A_2}{A_1}\right)^2 v_2^2 – v_2^2\right)\] | Substitute \(v_1 = \left(\frac{A_2}{A_1}\right)v_2\) so everything is in terms of \(v_2\). |
| \[P_2 – P_1 = \frac{1}{2}\rho\left(\left(\frac{A_2}{A_1}\right)^2 – 1\right)v_2^2\] | Factor out \(v_2^2\) to simplify the algebra. |
| \[v_2^2 = \frac{2(P_2 – P_1)}{\rho\left(\left(\frac{A_2}{A_1}\right)^2 – 1\right)}\] | Solve for \(v_2^2\). |
| \[\frac{A_2}{A_1} = \frac{542}{215} = 2.5209302326\] | Compute the area ratio (units cancel because both areas are in \(\text{cm}^2\)). Keep extra digits to reduce rounding error. |
| \[\left(\frac{A_2}{A_1}\right)^2 – 1 = (2.5209302326)^2 – 1 = 5.3530902024\] | Compute \(\left(\frac{A_2}{A_1}\right)^2 – 1\) accurately for the denominator. |
| \[v_2^2 = \frac{2(145)}{(1.35)(5.3530902024)} = 40.1025650823\] | Substitute \(P_2-P_1 = 145\ \text{Pa}\) and \(\rho = 1.35\ \text{kg/m}^3\) and evaluate \(v_2^2\). |
| \[v_2 = \sqrt{40.1025650823} = 6.3326543877\ \text{m/s}\] | Take the square root to get \(v_2\). |
| \[\boxed{v_2 \approx 6.33\ \text{m/s}}\] | Final answer to three significant figures (matching given data precision). |
| Derivation or Formula | Reasoning |
|---|---|
| \[A_1 v_1 = A_2 v_2\] | Use continuity again: same flow rate through both cross-sections. |
| \[v_1 = \left(\frac{A_2}{A_1}\right)v_2\] | Solve for \(v_1\). |
| \[v_1 = (2.5209302326)(6.3326543877) = 15.9598116104\ \text{m/s}\] | Substitute the computed ratio and the result from part (a). |
| \[\boxed{v_1 \approx 16.0\ \text{m/s}}\] | Round to three significant figures. |
| Derivation or Formula | Reasoning |
|---|---|
| \[Q = A_2 v_2\] | Volume flow rate is \(Q\), equal to area times speed at that section. |
| \[A_2 = 542\ \text{cm}^2 = 542\times 10^{-4}\ \text{m}^2 = 0.0542\ \text{m}^2\] | Convert \(\text{cm}^2\) to \(\text{m}^2\): \(1\ \text{cm}^2 = 10^{-4}\ \text{m}^2\). |
| \[Q = (0.0542)(6.3326543877) = 0.3434308678\ \text{m}^3/\text{s}\] | Multiply \(A_2\) (in \(\text{m}^2\)) by \(v_2\) (in \(\text{m/s}\)) to get \(Q\) in \(\text{m}^3/\text{s}\). |
| \[\boxed{Q \approx 0.343\ \text{m}^3/\text{s}}\] | Round to three significant figures. |
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Water flowing in a horizontal pipe speeds up as it goes from a section with a large diameter to a section with a small diameter. Which of the following can explain why the speed of the water increases?
Two blocks of the same size are floating in a container of water. The first block is submerged \( 80\% \) while the second block is submerged by \( 20\% \) beneath the water. Which of the following is a correct statement about the two blocks?
A trash compactor pushes down with a force of \( 500 \) \( \text{N} \) on a \( 3 \) \( \text{cm}^2 \) input piston, causing a force of \( 30,000 \) \( \text{N} \) to crush the trash. What is the area of the output piston that crushes the trash?
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?
The figure shows a horizontal pipe with sections with different cross-sectional areas. Small tubes extend from the top of each section. The cross-sectional area of the pipe at location C is half that at A, and the areas at A and D are the same. Water flows in the pipe from left to right. Which of the following correctly ranks the height \( h \) of the water in the tubes above the labeled locations?
Ben’s favorite ride at the Barrel-O-Fun Amusement Park 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 \( 3.0 \) \( \text{cm} \) wide cylindrical piston, which holds the \( 20,000 \) \( \text{N} \) ride off the ground. What is the diameter of the piston that holds the ride?
A geologist suspects that her rock specimen is hollow, so she weighs the specimen in both air and water. When completely submerged, the rock weighs twice as much in air as it does in water.
Three identical reservoirs, \(A\), \(B\), and \(C\), are represented above, each with a small pipe where water exits horizontally. The pipes are set at the same height above a pool of water. The water in the reservoirs is kept at the levels shown. Which of the following correctly ranks the horizontal distances \( d \) that the streams of water travel before hitting the surface of the pool?
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.
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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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