| 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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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, \( \dfrac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
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?
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 \( 2000 \) \( \text{cm}^2 \).
Which of the following statements is an expression of the equation of continuity?
Caleb is filling up water balloons for the Physics Olympics balloon toss competition. Caleb sets a \( 0.50 \text{-kg} \) spherical water balloon on the kitchen table and notices that the bottom of the balloon flattens until the pressure on the bottom is reduced to \( 630 \frac{\text{N}}{\text{m}^2} \). What is the area of the flat spot on the bottom of the balloon?
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?
Nancy is using a turkey baster (a kitchen tool with a rubber bulb on one end and a tube on the other) to collect juices from a roasting turkey. When she squeezes and then releases the rubber bulb, it creates suction with a pressure of \( 99{,}800 \) \( \text{Pa} \). This suction causes the turkey juice to rise \( 9 \) \( \text{cm} \) up the tube. Based on this information, what is the density of the turkey juice?
In the laboratory, you are given a cylindrical beaker containing a fluid and you are asked to determine the density \( \rho \) of the fluid. You are to use a spring of negligible mass and unknown spring constant \( k \) that is attached to a vertical stand.
A cube of unknown material and uniform density floats in a container of water with \(60\%\) of its volume submerged. If this same cube were placed in a container of oil with density \(800\) \(\text{kg/m}^3\), what portion of the cube’s volume would be submerged while floating?
A beaker weighing \( 2.0 \) \( \text{N} \) is filled with \( 5.0 \times 10^{-3} \) \( \text{m}^3 \) of water. A rubber ball weighing \( 3.0 \) \( \text{N} \) is held entirely underwater by a massless string attached to the bottom of the beaker, as represented in the figure above. The tension in the string is \( 4.0 \) \( \text{N} \). The water fills the beaker to a depth of \( 0.20 \) \( \text{m} \). Water has a density of \( 1000 \) \( \text{kg/m}^3 \). The effects of atmospheric pressure may be neglected.
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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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