| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_{BC} = \frac{\pi}{4} \times (0.6)^2 \] | Calculate the cross-sectional area of section BC using the diameter \( 600 \text{ mm} = 0.6 \text{ m} \). |
| 2 | \[ A_{BC} = 0.2827 \ \text{m}^2 \] | Evaluate the expression to get the area. |
| 3 | \[ Q_{BC} = A_{BC} \times v_{BC} \] | Use the formula for flow rate, \( Q = A \cdot v \). |
| 4 | \[ Q_{BC} = 0.2827 \times 1.2 \] | Substitute \( v_{BC} = 1.2 \ \text{m/s} \) into the equation. |
| 5 | \[ \boxed{Q_{BC} = 0.3393 \ \text{m}^3/\text{s}} \] | Calculate to find \( Q_{BC} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q_{AB} = Q_{BC} \] | Using the law of mass conservation, \( Q_{AB} = Q_{BC} \) since no other flows are present between A and C. |
| 2 | \[ \boxed{Q_{AB} = 0.3393 \ \text{m}^3/\text{s}} \] | \( Q_{BC} \) was calculated earlier as \( 0.3393 \ \text{m}^3/\text{s} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_{AB} = \frac{\pi}{4} \times (0.3)^2 \] | Calculate the cross-sectional area of section AB using \( 300 \text{ mm} = 0.3 \text{ m} \). |
| 2 | \[ A_{AB} = 0.0707 \ \text{m}^2 \] | Evaluate the expression for area. |
| 3 | \[ v_{AB} = \frac{Q_{AB}}{A_{AB}} \] | Rearrange the formula \( Q = A \cdot v \) to solve for \( v \). |
| 4 | \[ v_{AB} = \frac{0.3393}{0.0707} \] | Substitute \( Q_{AB} \) and \( A_{AB} \) into the equation. |
| 5 | \[ \boxed{v_{AB} = 4.8 \ \text{m/s}} \] | Evaluate to find \( v_{AB} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q_{CD} = \frac{Q_{AB}}{1.5} \] | From \( Q_{AB} = Q_{CD} + Q_{CE} \), solve for \( Q_{CD} \), knowing that \( Q_{CE} = .5Q_{CD} \). |
| 2 | \[ Q_{CD} = \frac{0.3393}{1.5} \] | Substitute \( Q_{AB} = 0.3393 \ \text{m}^3/\text{s} \). |
| 3 | \[ Q_{CD} = 0.2262 \ \text{m}^3/\text{s} \] | Calculate to find \( Q_{CD} \). |
| 4 | \[ A_{CD} = \frac{Q_{CD}}{v_{CD}} \] | Rearrange \( Q = A \cdot v \) to solve for \( A \). |
| 5 | \[ A_{CD} = \frac{0.2262}{1.4} \] | Substitute \( Q_{CD} \) and \( v_{CD} = 1.4 \ \text{m/s} \). |
| 6 | \[ A_{CD} = 0.1616 \ \text{m}^2 \] | Evaluation to find \( A_{CD} \). |
| 7 | \[ d_{CD} = \sqrt{\frac{4 \times A_{CD}}{\pi}} \] | Calculate the diameter from the area. |
| 8 | \[ d_{CD} = \sqrt{\frac{4 \times 0.1616}{\pi}} \] | Substitute \( A_{CD} \) into the equation. |
| 9 | \[ \boxed{d_{CD} = 0.454 \ \text{m}} \] | Calculate the diameter \( d_{CD} \). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q_{CE} = 0.5Q_{CD} \] | From given condition \( Q_{CE} = 0.5Q_{CD} \). |
| 2 | \[ Q_{CE} = 0.5 \times 0.2262 \] | Use previously calculated \( Q_{CD} \). |
| 3 | \[ Q_{CE} = 0.1131 \ \text{m}^3/\text{s} \] | Evaluate to find \( Q_{CE} \). |
| 4 | \[ A_{CE} = \frac{\pi}{4} \times (0.15)^2 \] | Calculate \( A_{CE} \) with \( 150 \text{ mm} = 0.15 \text{ m} \). |
| 5 | \[ A_{CE} = 0.0177 \ \text{m}^2 \] | Evaluate for \( A_{CE} \). |
| 6 | \[ v_{CE} = \frac{Q_{CE}}{A_{CE}} \] | Rearrange \( Q = A \cdot v \) to solve for \( v \). |
| 7 | \[ v_{CE} = \frac{0.1131}{0.0177} \] | Substitute \( Q_{CE} \) and \( A_{CE} \). |
| 8 | \[ \boxed{v_{CE} = 6.4 \ \text{m/s}} \] | Calculate \( v_{CE} \). |
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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.
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?
Two points, \( A \) and \( B \), are in a pipe carrying a flowing ideal fluid. Point \( B \) is \( 2.0 \) \( \text{m} \) higher than point \( A \), and the fluid speed at \( B \) is twice the speed at \( A \). If the pressure at \( A \) is \( P_A \), which of the following expressions correctly represents the pressure at \( B \) \( (P_B) \)?
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?
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?
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.
Johnny the auto mechanic is raising a \( 1200 \) \( \text{kg} \) car on her hydraulic lift so that she can work underneath. If the area of the input piston is \( 12 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
A drinking fountain projects water at an initial angle of \( 50^ \circ \) above the horizontal, and the water reaches a maximum height of \( 0.150 \) \( \text{m} \) above the point of exit. Assume air resistance is negligible.
A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
Rex, an auto mechanic, is raising a \( 1200 \) \( \text{kg} \) car on his hydraulic lift so that he can work underneath. If the area of the input piston is \( 12.0 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
\( Q_{AB} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( v_{AB} = 4.8 \) \( \text{m/s} \)
\( Q_{BC} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( d_{CD} = 0.454 \) \( \text{m} \) \( = 454 \) \( \text{mm} \)
\( v_{CE} = 6.4 \) \( \text{m/s} \)
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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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