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. |
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A cube of side length \( s \) rests on the bottom surface of a container of fluid. The fluid is at a height \( y \) above the bottom of the tank. The fluid has density \( \rho \) and the atmospheric pressure is \( P_{\text{atm}} \).
Which of the following expressions is equal to the absolute pressure exerted by the fluid on the top surface of the cube?
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?
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?
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 cube of unknown material and uniform density floats in a container of water with \(60\%\) of its volume submerged. If this same cube were placed in a container of oil with density \(800\) \(\text{kg/m}^3\), what portion of the cube’s volume would be submerged while floating?
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?
In the laboratory, you are given a cylindrical beaker containing a fluid and you are asked to determine the density \( \rho \) of the fluid. You are to use a spring of negligible mass and unknown spring constant \( k \) that is attached to a vertical stand.
Alcohol has a specific gravity of \( 0.79 \). If a barometer consisting of an open-ended tube placed in a dish of alcohol is used at sea level, to what height in the tube will the alcohol rise?
A Venturi tube has a pressure difference of \( 15\,000 \) \( \text{Pa} \). The entrance radius is \( 3 \) \( \text{cm} \), while the exit radius is \( 1 \) \( \text{cm} \). What are the entrance velocity, exit velocity, and flow rate if the fluid is gasoline \( (\rho = 700 \) \( \text{kg/m}^3 ) \)?
Three identical reservoirs, \(A\), \(B\), and \(C\), are represented above, each with a small pipe where water exits horizontally. The pipes are set at the same height above a pool of water. The water in the reservoirs is kept at the levels shown. Which of the following correctly ranks the horizontal distances \( d \) that the streams of water travel before hitting the surface of the pool?
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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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