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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Water flowing in a horizontal pipe speeds up as it goes from a section with a large diameter to a section with a small diameter. Which of the following can explain why the speed of the water increases?
A solid titanium sphere of radius \( 0.35 \) \( \text{m} \) has a density \( 4500 \) \( \text{kg/m}^3 \). It is held suspended completely underwater by a cable. What is the tension in the cable?
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?
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?
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 \( P_2 – P_1 = 145 \) \( \text{Pa} \).
Find the approximate minimum mass needed for a spherical ball with a \(40\) \(\text{cm}\) radius to sink in a liquid of density \(1.4 \times 10^3\) \(\text{kg/m}^3\). Use \(9.8 \text{m/s}^2\) for \(g\).
A liquid flows at a constant flow rate through a pipe with circular cross-sections of varying diameters. At one point in the pipe, the diameter is \(2\) \(\text{cm}\) and the flow speed is \(18\) \(\text{m/s}\). What is the flow speed at another point in this pipe, where the diameter is \(3\) \(\text{cm}\).
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?
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.
The figure shows a container filled with water to a depth \( d \). The container has a hole a distance \( y \) above its bottom, allowing water to exit with an initially horizontal velocity. Which of the following correctly predicts and explains how the speed of the water as it exits the hole would change if the distance \( y \) above the bottom of the container increased?
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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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