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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_1 = \frac{\pi d_1^2}{4} \quad , \quad A_2 = \frac{\pi d_2^2}{4} \] | Calculate the cross-sectional areas of the pipe at street level and top floor using the area formula for a circle. Here, \(d_1 = 0.05\,m\) and \(d_2 = 0.028\,m\). |
| 2 | \[ A_1 v_i = A_2 v_x \] | Apply the continuity equation for incompressible flow which states that the volumetric flow rate must remain constant. |
| 3 | \[ v_x = \frac{A_1}{A_2} v_i = \left(\frac{d_1}{d_2}\right)^2 v_i \] | Simplify the expression by canceling the common factor \(\frac{\pi}{4}\) in the area formulas, showing that the velocity scales as the square of the diameter ratio. |
| 4 | \[ v_x = \left(\frac{0.05}{0.028}\right)^2 (0.78) \approx 2.48\,m/s \] | Substitute the given values: \(d_1 = 0.05\,m\), \(d_2 = 0.028\,m\), and \(v_i = 0.78\,m/s\). The ratio \(\left(\frac{0.05}{0.028}\right)^2 \) is approximately 3.188; multiplying by 0.78 yields \(v_x \approx 2.48\,m/s\). |
| 5 | \[ \boxed{v_x \approx 2.48\, m/s} \] | This is the final computed flow velocity at the top floor of the building. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \frac{P_1}{\rho} + \frac{v_i^2}{2} + g y_1 = \frac{P_2}{\rho} + \frac{v_x^2}{2} + g y_2 \] | Write Bernoulli’s equation between the street level (point 1) and the top floor (point 2). Here, \(y_1 = 0\) and \(y_2 = 16\,m\). |
| 2 | \[ \frac{P_2}{\rho} = \frac{P_1}{\rho} + \frac{v_i^2 – v_x^2}{2} – g \Delta y \] | Rearrange the equation to solve for \(P_2\), where \(\Delta y = y_2 – y_1 = 16\,m\). |
| 3 | \[ P_2 = P_1 + \rho \left(\frac{v_i^2 – v_x^2}{2} – g \Delta y \right) \] | Multiply both sides by the density \(\rho\) to isolate \(P_2\). |
| 4 | \[ P_1 = 3.8\,atm = 3.8 \times 101325 \approx 385035\,Pa \] | Convert the gauge pressure at street level from atmospheres to Pascals using \(1\,atm \approx 101325\,Pa\). |
| 5 | \[ \frac{v_i^2 – v_x^2}{2} = \frac{0.78^2 – 2.48^2}{2} \approx \frac{0.6084 – 6.1504}{2} \approx -2.77\, m^2/s^2 \] | Calculate the difference in kinetic energy per unit mass between the two points. |
| 6 | \[ g \Delta y = 9.8 \times 16 \approx 156.8\, m^2/s^2 \] | Compute the gravitational potential energy change per unit mass over a height of \(16\,m\). |
| 7 | \[ \rho \left( \frac{v_i^2 – v_x^2}{2} – g \Delta y \right) \approx 1000 \left(-2.77 – 156.8\right) \approx -159570\,Pa \] | Combine the kinetic and potential terms multiplied by the density \(\rho = 1000\,kg/m^3\) to find the pressure change. |
| 8 | \[ P_2 \approx 385035 – 159570 \approx 225465\,Pa \] | Subtract the pressure drop from the initial gauge pressure to get the gauge pressure at the top floor. |
| 9 | \[ \boxed{P_2 \approx 2.25 \times 10^5\,Pa} \] | This is the final gauge pressure of the water in the pipe on the top floor. |
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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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