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| 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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The figure above shows a portion of a conduit for water, one with rectangular cross sections. If the flow speed at the top is \( v \), what is the flow speed at the bottom?
The drawing above shows a spherical reservoir that contains \( 455,000 \) \( \text{kg} \) of water when full. The reservoir is vented to the atmosphere at the top. Assuming the reservoir is full and the diameter of the reservoir is much larger than any of the pipes on the ground.
Two objects labeled K and L have equal mass but densities \( 0.95D_o \) and \( D_o \), respectively. Each of these objects floats after being thrown into a deep swimming pool. Which is true about the buoyant forces acting on these objects?
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
A diver descends from a salvage ship to the ocean floor at a depth of \(35 \text{ m}\) below the surface. The density of ocean water is \(1.025 \times 10^3 \text{ kg/m}^3\).
Which of the following statements is an expression of the equation of continuity?
Diamond has a density of \( 3500 \) \( \text{kg/m}^3 \). During a physics lab, a diamond drops out of Virginia’s necklace and falls into her graduated cylinder filled with \( 5.00 \times 10^{-5} \) \( \text{m}^3 \) of water. This causes the water level to rise to the \( 5.05 \times 10^{-5} \) \( \text{m}^3 \) mark. What is the mass of Virginia’s diamond?
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 cube of side length \( s \) rests on the bottom surface of a container of fluid. The fluid is at a height \( y \) above the bottom of the tank. The fluid has density \( \rho \) and the atmospheric pressure is \( P_{\text{atm}} \).
Which of the following expressions is equal to the absolute pressure exerted by the fluid on the top surface of the cube?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
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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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