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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The \( 70 \) \( \text{kg} \) student in the figure balances a \( 1200 \) \( \text{kg} \) elephant on a hydraulic lift. Assume that it is filled with oil, which is incompressible and has a density \( \rho = 900 \) \( \text{kg/m}^3 \). What is the diameter of the piston the student is standing on? Assume each piston has a cylindrical shape, i.e., a circular cross-sectional area. Note: The two pistons are at the same height. Also, the diameter of the wider piston is given in the figure to be \( 2.0 \) \( \text{m} \).
The radius of the left piston is \( 0.12 \) \( \text{m} \) and the radius of the right piston is \( 0.65 \) \( \text{m} \). If \( f \) were raised by \( 14 \) \( \text{N} \), how much would \( F \) need to be increased to maintain equilibrium?
A horizontal tube with two vertical T-branches (A and B) is partially submerged in a liquid, with the open ends of the branches exposed to the air. However, the section of the tube above point B is hidden from view and may either be wider or narrower than the section above A.
Air is blown through the horizontal tube, causing the liquid levels in the vertical branches to rise as shown. Based on the observed water levels, which of the following best describes the characteristics of the hidden section of the tube above B?
A sample of an unknown material appears to weigh \( 285 \) \( \text{N} \) in air and \( 195 \) \( \text{N} \) when immersed in alcohol of specific gravity \( 0.700 \).
Nancy is using a turkey baster (a kitchen tool with a rubber bulb on one end and a tube on the other) to collect juices from a roasting turkey. When she squeezes and then releases the rubber bulb, it creates suction with a pressure of \( 99{,}800 \) \( \text{Pa} \). This suction causes the turkey juice to rise \( 9 \) \( \text{cm} \) up the tube. Based on this information, what is the density of the turkey juice?
Johnny the auto mechanic is raising a \( 1200 \) \( \text{kg} \) car on her hydraulic lift so that she can work underneath. If the area of the input piston is \( 12 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
A block of weight \( W \) is floating in water, and one-third of the block is above the surface of the water. Which of the following correctly describes the magnitude \( F \) of the force that the block exerts on the water and explains why \( F \) has that value?
A Venturi meter is a device used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed \( v_2 \) through a horizontal section of pipe with a cross-sectional area \( A_2 = 542 \) \( \text{cm}^2 \). The gas has a density of \( 1.35 \) \( \text{kg/m}^3 \). The Venturi meter has a cross-sectional area of \( A_1 = 215 \) \( \text{cm}^2 \) and has been substituted for a section of the larger pipe. The pressure difference between the two sections \( P_2 – P_1 = 145 \) \( \text{Pa} \).
Which of the following statements is an expression of the equation of continuity?
Rex, an auto mechanic, is raising a \( 1200 \) \( \text{kg} \) car on his hydraulic lift so that he can work underneath. If the area of the input piston is \( 12.0 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
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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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