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Part A
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ Q = \pi r^2 v \] | This is the expression for the volume rate of flow, where \(r\) is the nozzle radius and \(v\) is the exit velocity. |
| 2 | \[ Q = \pi (0.015)^2 (6.0) \] | Substitute the given values \(r = 0.015\,\text{m}\) and \(v = 6.0\,\text{m/s}\) into the equation. |
| 3 | \[ Q \approx \pi \times 0.000225 \times 6.0 \approx \pi \times 0.00135 \approx 0.00424\,\text{m}^3/\text{s} \] | Perform the multiplication and use an approximation for \(\pi\) to calculate \(Q\). |
| 4 | \[ \boxed{Q \approx 0.00424\,\text{m}^3/\text{s}} \] | This is the final numerical value for the volume rate of flow. |
Part B
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_{\text{pipe}} = \pi r_{\text{pipe}}^2 = \pi (0.025)^2 \] | Calculate the cross-sectional area of the pipe having radius \(0.025\,\text{m}\). |
| 2 | \[ A_{\text{pipe}} \approx \pi \times 0.000625 \approx 0.00196\,\text{m}^2 \] | Multiply to find the numerical area. |
| 3 | \[ v_{\text{pipe}} = \frac{Q}{A_{\text{pipe}}} = \frac{0.00424}{0.00196} \approx 2.16\,\text{m/s} \] | Determine the velocity of water in the pipe using the constant flow rate \(Q\) from part (a). |
| 4 | \[ P_{\text{pipe}} = P_{\text{atm}} + \frac{1}{2}\rho\left(v_{\text{exit}}^2 – v_{\text{pipe}}^2\right) + \rho g (2.5) \] | Apply Bernoulli’s equation between the fountain exit (where \(P_{\text{exit}} = P_{\text{atm}}\) and \(v_{\text{exit}} = 6.0\,\text{m/s}\)) at \(z=0\) and the pipe point, which is \(2.5\,\text{m}\) below. |
| 5 | \[ \frac{1}{2}\rho(v_{\text{exit}}^2 – v_{\text{pipe}}^2) = \frac{1}{2}(1000)(36 – 4.67) \approx 500 \times 31.33 \approx 15665\,\text{Pa} \] | Compute the kinetic term using \(\rho = 1000\,\text{kg/m}^3\), \(v_{\text{exit}}^2 = 36\), and \(v_{\text{pipe}}^2 \approx 4.67\). |
| 6 | \[ \rho g (2.5) = 1000 \times 9.81 \times 2.5 \approx 24525\,\text{Pa} \] | Calculate the gravitational pressure increase due to the \(2.5\,\text{m}\) height difference. |
| 7 | \[ P_{\text{pipe}} \approx 101325 + 15665 + 24525 \approx 141515\,\text{Pa} \] | Sum up the atmospheric pressure with the kinetic and gravitational contributions. |
| 8 | \[ \boxed{P_{\text{pipe}} \approx 1.42 \times 10^5\,\text{Pa}} \] | This is the final calculated absolute pressure in the pipe. |
Part C
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ v_{\text{new}} = \sqrt{2 g h} \] | To reach a maximum height \(h\), the required exit speed is given by equating kinetic energy to gravitational potential energy. |
| 2 | \[ v_{\text{new}} = \sqrt{2 \times 9.81 \times 4.0} \approx \sqrt{78.48} \approx 8.86\,\text{m/s} \] | Substitute \(g = 9.81\,\text{m/s}^2\) and \(h = 4.0\,\text{m}\) to compute the new exit velocity. |
| 3 | \[ A_{\text{new}} = \frac{Q}{v_{\text{new}}} = \frac{0.00424}{8.86} \approx 0.000478\,\text{m}^2 \] | With the flow rate constant, the new nozzle area is determined by dividing \(Q\) by the new exit velocity. |
| 4 | \[ r_{\text{new}} = \sqrt{\frac{A_{\text{new}}}{\pi}} = \sqrt{\frac{0.000478}{\pi}} \approx 0.0123\,\text{m} \] | Calculate the new nozzle radius from the area using the area formula of a circle. |
| 5 | \[ \boxed{r_{\text{new}} \approx 0.0123\,\text{m}} \] | This is the required radius of the nozzle to launch the water to \(4.0\,\text{m}\) while maintaining the same flow rate. |
Just ask: "Help me solve this problem."
In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
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?
Water flows from point \( A \) to points \( D \) and \( E \) as shown. Some of the flow parameters are known, as shown in the table. Determine the unknown parameters. Note the diagram above does not show the relative diameters of each section of the pipe.
| Section | Diameter | Flow Rate | Velocity |
|---|---|---|---|
| \( \text{AB} \) | \( 300 \) \( \text{mm} \) | \(\textbf{?}\) | \(\textbf{?}\) |
| \( \text{BC} \) | \( 600 \) \( \text{mm} \) | \(\textbf{?}\) | \( 1.2 \) \( \text{m/s} \) |
| \( \text{CD} \) | \(\textbf{?}\) | \( Q_{CD} = 2Q_{CE} \) \( \text{m}^3/\text{s} \) | \( 1.4 \) \( \text{m/s} \) |
| \( \text{CE} \) | \( 150 \) \( \text{mm} \) | \( Q_{CE} = 0.5Q_{CD} \) \( \text{m}^3/\text{s} \) | \(\textbf{?}\) |
Two paper cups are suspended by strings and hung near each other. They are separated by about \( 10 \) \( \text{cm} \). Explain what happens to the cups when you blow air between them. Hint: Do they remain still, moves away from each other or move towards each other?
Caleb is filling up water balloons for the Physics Olympics balloon toss competition. Caleb sets a \( 0.50 \text{-kg} \) spherical water balloon on the kitchen table and notices that the bottom of the balloon flattens until the pressure on the bottom is reduced to \( 630 \frac{\text{N}}{\text{m}^2} \). What is the area of the flat spot on the bottom of the balloon?
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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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