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| Derivation or Formula | Reasoning |
|---|---|
| \[A_1 v_i = A_2 v_x\] | Conservation of mass (continuity equation) requires that the volumetric flow rate remains constant, where \(A_1\) and \(A_2\) are the cross-sectional areas of the basement and second floor pipes, respectively. |
| \[A = \frac{\pi}{4}d^2\] | This is the formula for the area of a circle, with \(d\) as the diameter of the pipe. We use it to express \(A_1\) and \(A_2\) in terms of the given diameters. |
| \[v_x = \left(\frac{A_1}{A_2}\right)v_i = \left(\frac{d_1}{d_2}\right)^2 v_i\] | Substituting the area formula into the continuity equation, the constant factors cancel and we obtain \(v_x\) in terms of the diameters \(d_1\) and \(d_2\) and the initial speed \(v_i\). |
| \[v_x = \left(\frac{0.02}{0.013}\right)^2 (0.5)\] | Substitute \(d_1 = 0.02\,\text{m}\), \(d_2 = 0.013\,\text{m}\), and \(v_i = 0.5\,\text{m/s}\) into the equation. |
| \[v_x \approx (1.5385)^2 \times 0.5 \approx 2.367 \times 0.5 \approx 1.18\,\text{m/s}\] | Evaluating the ratio gives approximately \(1.5385\) which squared is about \(2.367\). Multiplying by \(0.5\,\text{m/s}\) results in a speed of approximately \(1.18\,\text{m/s}\) in the second floor pipe. |
| \[\boxed{v_x \approx 1.18\,\text{m/s}}\] | This is the final expression for the flow speed in the 1.3 cm diameter pipe on the second floor. |
| Derivation or Formula | Reasoning |
|---|---|
| \[P_1 + \frac{1}{2}\rho v_i^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_x^2 + \rho g h_2\] | This is Bernoulli’s equation applied between the basement (point 1) and the second floor (point 2). It relates pressure, kinetic, and potential energy per unit volume along a streamline. |
| \[h_1 = 0,\quad h_2 = 5\,\text{m}\] | We set the basement as the reference level (\(h_1 = 0\)) so that the second floor is at an elevation of \(5\,\text{m}\). |
| \[P_1 = 3\,\text{atm} = 3 \times 101325 = 303975\,\text{Pa}\] | The initial pressure is given as \(3\,\text{atm}\); converting to Pascals using \(1\,\text{atm} = 101325\,\text{Pa}\) yields \(303975\,\text{Pa}\). |
| \[P_2 = P_1 + \frac{1}{2}\rho (v_i^2 – v_x^2) – \rho g (h_2 – h_1)\] | Rearrange Bernoulli’s equation to solve for \(P_2\), the pressure in the second floor pipe. |
| \[P_2 = 303975 + \frac{1}{2}(1000)(0.5^2 – 1.18^2) – (1000)(9.8)(5)\] | Substitute \(\rho = 1000\,\text{kg/m}^3\) for water, \(g = 9.8\,\text{m/s}^2\), \(v_i = 0.5\,\text{m/s}\), and \(v_x \approx 1.18\,\text{m/s}\) into the equation. |
| \[0.5^2 = 0.25,\quad 1.18^2 \approx 1.392\] | Calculate the squares of the velocities for the kinetic energy terms. |
| \[\frac{1}{2}\times 1000\times(0.25 – 1.392) \approx 500\times(-1.142) \approx -571\,\text{Pa}\] | The kinetic energy difference contributes a change of approximately \(-571\,\text{Pa}\) (negative since the speed increases in the smaller pipe). |
| \[1000 \times 9.8 \times 5 = 49000\,\text{Pa}\] | This term accounts for the increase in gravitational potential energy due to the 5 m elevation gain. |
| \[P_2 = 303975 – 571 – 49000 \approx 254404\,\text{Pa}\] | Subtract the kinetic and potential energy contributions from the initial pressure to find the pressure at the second floor. |
| \[\boxed{P_2 \approx 254400\,\text{Pa}}\] | This is the final pressure in the 1.3 cm diameter pipe on the second floor. In atmospheres, this is roughly \(254400/101325 \approx 2.51\,\text{atm}\). |
Just ask: "Help me solve this problem."
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?
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 sphere of mass \(0.5\) \(\text{kg}\) is dropped into a column of oil. At the instant the sphere becomes completely submerged in the oil, the sphere is moving downward at \(8\) \(\text{m/s}\), the buoyancy force on the sphere is \(4.0\) \(\text{N}\), and the fluid frictional force is \(4.0\) \(\text{N}\). Which of the following describes the motion of the sphere at this instant?
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?
The diagram above shows a hydraulic chamber with a spring \( (k_s = 1250 \, \text{N/m}) \) attached to the input piston and a rock of mass \( 55.2 \, \text{kg} \) resting on the output plunger. The input piston and output plunger are at about the same height, and each has negligible mass. The chamber is filled with water.
\(1.18\,\text{m/s}\)\n\(254400\,\text{Pa}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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