| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(A_1 v_1 = A_2 v_2\) | Use the principle of continuity for fluid flow, which states that the product of cross-sectional area and flow speed must remain constant for an incompressible fluid in a tube with varying diameter. |
| 2 | \(A = \pi \left(\frac{d}{2}\right)^2\) | Calculate the cross-sectional area of a circular section of the pipe using the diameter \(d\). |
| 3 | \(A_1 = \pi \left(\frac{2}{2}\right)^2 = \pi (1)^2 = \pi\) \(\text{cm}^2\) | Substitute \(d_1 = 2\) cm into the area formula to find the cross-sectional area at the first point. |
| 4 | \(A_2 = \pi \left(\frac{3}{2}\right)^2 = \pi \left(\frac{9}{4}\right)\) \(\text{cm}^2\) | Substitute \(d_2 = 3\) cm into the area formula to find the cross-sectional area at the second point. |
| 5 | \(\pi v_1 = \frac{\pi (9/4)}{\pi} v_2\) | Substitute the cross-sectional areas into the continuity equation. The factor of \(\pi\) cancels out on both sides. |
| 6 | \(v_2 = v_1 \frac{4}{9}\) | Solve for \(v_2\) by isolating it on one side of the equation. |
| 7 | \(v_2 = 18 \times \frac{4}{9} = 8\) \(\text{m/s}\) | Substitute \(v_1 = 18\) \(\text{m/s}\) into the equation to calculate \(v_2\). |
| 8 | \(v_2 = 8\) \(\text{m/s}\) | The flow speed at the second point is \(8\) \(\text{m/s}\). |
The correct answer is (c) \( 8 \text{ m/s} \).
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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 lab, a student is given a glass beaker filled with water with an ice cube of mass \( m \) and volume \( V_c \) floating in it.
The downward force of gravity on the ice cube has magnitude \( F_g \). The student pushes down on the ice cube with a force of magnitude \( F_P \) so that the cube is totally submerged. The water then exerts an upward buoyant force on the ice cube of magnitude \( F_B \). Which of the following is an expression for the magnitude of the acceleration of the ice cube when it is released?
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 \( (\rho = 700 \) \( \text{kg/m}^3 ) \)?
The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is \( 2.5 \) \( \text{m} \), and the puncture is \( 1 \) \( \text{m} \) above the ground level, what is the efflux speed of the water?
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 \( B \) \( (P_B) \)?
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 geologist suspects that her rock specimen is hollow, so she weighs the specimen in both air and water. When completely submerged, the rock weighs twice as much in air as it does in water.
In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
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?
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.
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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