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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. |
Just ask: "Help me solve this problem."
A car with speed \( v \) and an identical car with speed \( 3v \) both travel the same circular section of an unbanked (flat) road. If the frictional force required to keep the faster car on the road without skidding is \( F \), then the frictional force required to keep the slower car on the road without skidding is
Two identical blocks are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity \( v \) to the right. The spring is then released by remote control. At some later instant, the left block is moving at \( \frac{v}{2} \) to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant?
Two blocks are on a horizontal, frictionless surface. Block \( A \) is moving with an initial velocity of \( v_0 \) toward block \( B \), which is stationary. The two blocks collide, stick together, and move off with a velocity of \( \frac{v_0}{3} \). Which block, if either, has the greater mass?
A pendulum with a period of \( 1 \) \( \text{s} \) on Earth, where the acceleration due to gravity is \( g \), is taken to another planet, where its period is \( 2 \) \( \text{s} \). The acceleration due to gravity on the other planet is most nearly
A centripetal force of \( 5.0 \) newtons is applied to a rubber stopper moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed, \( v \), and the frequency, \( f \), of the stopper?
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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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