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| 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 \( v_{CE} \). |
Just ask: "Help me solve this problem."
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 \( 2000 \) \( \text{cm}^2 \).
When the button of a trash compactor is pushed, a force of \( 350 \) \( \text{N} \) pushes down on a \( 1.3 \) \( \text{cm}^2 \) input piston, creating a force of \( 22,076 \) \( \text{N} \) to crush the trash. What is the area of the piston that crushes the trash?
A pump, submerged at the bottom of a well that is \( 35 \) \( \text{m} \) deep, is used to pump water uphill to a house that is \( 50 \) \( \text{m} \) above the top of the well, as shown to the right. The density of water is \( 1000 \) \( \text{kg/m}^3 \). All pressures are gauge pressures. Neglect the effects of friction, turbulence, and viscosity.
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?
Wanda watches the fish in her fish tank and notices that the angelfish like to feed at the water’s surface, while the catfish feed \( 0.300 \) \( \text{m} \) below at the bottom of the tank. If the average density of the water in the tank is \( 1000. \) \( \text{kg/m}^3 \), what is the pressure on the catfish?
\( Q_{AB} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( v_{AB} = 4.8 \) \( \text{m/s} \)
\( Q_{BC} = 0.3393 \) \( \text{m}^3/\text{s} \)
\( d_{CD} = 0.454 \) \( \text{m} \) \( = 454 \) \( \text{mm} \)
\( v_{CE} = 6.4 \) \( \text{m/s} \)
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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] |
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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