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To analyze this scenario, we will use the principles of fluid dynamics, specifically Bernoulli’s principle, which relates the speed of a fluid to its pressure:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}\) | Apply Bernoulli’s principle, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline. |
| 2 | \(\Delta h = h_B – h_A\) | Identify that the liquid height difference (\(\Delta h\)) in the vertical branches indicates a difference in pressure between points A and B. |
| 3 | \(P_A + \rho g h_A = P_B + \rho g h_B\) | The liquid column heights in branches A and B equilibrate the pressure differences, where \(P_A\) and \(P_B\) are the pressures at points A and B. |
| 4 | \(\Delta P = \rho g \Delta h\) | Express the difference in pressure (\(\Delta P\)) in terms of the height difference. This implies that a lower pressure exists where the liquid height is greater. |
| 5 | \(\frac{1}{2} \rho v_A^2 + P_A = \frac{1}{2} \rho v_B^2 + P_B\) | Since the tube is horizontal, apply Bernoulli’s equation for points A and B, considering equal height, where \(v_A\) and \(v_B\) are velocities at A and B. |
| 6 | \(P_B < P_A \) implies \(v_B > v_A\) | The presence of a higher liquid level at B suggests a lower pressure at B, indicating that the air speed (\(v_B\)) above B is greater. |
| 7 | \(\text{Narrower tube} \Rightarrow \text{higher speed (continuity)}\) | The continuity equation implies \(A_A v_A = A_B v_B\). If \(v_B > v_A\), then \(A_B < A_A\), indicating the section is narrower. |
| 8 | (c) The tube is narrower, and the air speed is greater above section B. | The correct description is a narrower section at B with greater speed due to the observed lower pressure (higher liquid level). |
Answer: (c) The tube is narrower, and the air speed is greater above section B.
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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.
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 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 \(1.025 \times 10^3 \text{ kg/m}^3\).
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 \(1025\) \(\text{kg/m}^3\).
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 \(g\).
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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] |
General Metric Conversion Chart
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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