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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You have a giant cask of water with a spigot some height below the water surface. The surface of the water, which is essentially at rest, is exposed to atmosphere (\( \approx 10^5 \text{Pa} \)). The water density is \( \approx 1000 \text{kg/m}^3 \). The water pours out of the spigot at \( 3 \text{m/s} \). How far below the water surface is the spigot positioned?
A helium-filled balloon is attached by a string of negligible mass to a small \(0.015 \ \text{kg}\) object that is just heavy enough to keep the balloon from rising. The total mass of the balloon, including the helium, is \(0.0050 \ \text{kg}\). The density of air is \(\rho_{\text{air}} = 1.29 \ \text{kg/m}^3\), and the density of helium is \(\rho_{\text{He}} = 0.179 \ \text{kg/m}^3\). The buoyant force on the \(0.015 \ \text{kg}\) object is small enough to be negligible.
A diver descends from a salvage ship to the ocean floor at a depth of \(35 \text{ m}\) below the surface. The density of ocean water is \(1.025 \times 10^3 \text{ kg/m}^3\).
Marc’s favorite ride at Busch Gardens is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a force of \( 72 \) \( \text{N} \) to a \( 30 \)\( \text{cm}^2 \) cylindrical piston, which holds the \( 20,000 \)\( \text{N} \) ride off the ground. What is the area of the piston that holds the ride?
The experimental diving rig is lowered from rest at the ocean’s surface and reaches a maximum depth of \(80\) \(\text{m}\). Initially it accelerates downward at a rate of \(0.10\) \(\text{m/s}^2\) until it reaches a speed of \(2.0\) \(\text{m/s}\), which then remains constant. During the descent, the pressure inside the bell remains constant at \(1\) atmosphere. The top of the bell has a cross-sectional area \(A = 9.0\) \(\text{m}^2\). The density of seawater is \(1025\) \(\text{kg/m}^3\).
Two points, \( A \) and \( B \), are in a pipe carrying a flowing ideal fluid. Point \( B \) is \( 2.0 \) \( \text{m} \) higher than point \( A \), and the fluid speed at \( B \) is twice the speed at \( A \). If the pressure at \( A \) is \( P_A \), which of the following expressions correctly represents the pressure at \( B \) \( (P_B) \)?
A sphere of mass \(0.5\) \(\text{kg}\) is dropped into a column of oil. At the instant the sphere becomes completely submerged in the oil, the sphere is moving downward at \(8\) \(\text{m/s}\), the buoyancy force on the sphere is \(4.0\) \(\text{N}\), and the fluid frictional force is \(4.0\) \(\text{N}\). Which of the following describes the motion of the sphere at this instant?
The diagram above shows a hydraulic chamber with a spring \( (k_s = 1250 \, \text{N/m}) \) attached to the input piston and a rock of mass \( 55.2 \, \text{kg} \) resting on the output plunger. The input piston and output plunger are at about the same height, and each has negligible mass. The chamber is filled with water.
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 \( P_2 – P_1 = 145 \) \( \text{Pa} \).
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?
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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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