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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Continuity Equation: \( A_1v_1 = A_2v_2 \) | The principle of conservation of mass states that the product of cross-sectional area (\(A\)) and velocity (\(v\)) must remain constant in a stream of incompressible fluid, meaning the speed of the fluid can change when the cross-sectional area changes. |
| 2 | Bernoulli’s Equation: \( P_1 + \frac{1}{2}\rho v_1^2 + \rho gh = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh \) | Since the pipe is horizontal \( (\rho gh) \) terms cancel out. Therefore, if the cross-sectional area decreases (causing \(v_2 > v_1\)), the pressure in the smaller pipe section must decrease to conserve energy. |
| 3 | Comparison: \( P_1 > P_2 \) | From Bernoulli’s equation, it is clear that when the velocity of a fluid increases, the pressure decreases. This indicates that the pressure in the larger section of the pipe is higher than in the smaller section. |
| 4 | Work Done Explanation | When fluid moves from a high-pressure region to a low-pressure region, the higher pressure does work on the fluid, increasing its kinetic energy and hence its speed. |
| 5 | Choice: (c) Work is done because the water in the larger pipe has a higher pressure. | This is the correct explanation. The water speeds up because there is a higher pressure in the large-diameter section of the pipe, doing work on the water as it moves to the lower-pressure, smaller-diameter section. |
Explanation of Incorrect Choices:
– (a) and (b) are incorrect. The system is horizontal, so gravitational potential energy is not a factor.
– (d) is incorrect because pressure is not lower in the larger section of the pipe.
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The figure above shows a portion of a conduit for water, one with rectangular cross sections. If the flow speed at the top is \( v \), what is the flow speed at the bottom?
A drinking fountain projects water at an initial angle of \( 50^ \circ \) above the horizontal, and the water reaches a maximum height of \( 0.150 \) \( \text{m} \) above the point of exit. Assume air resistance is negligible.
A pump, submerged at the bottom of a well that is \( 35 \) \( \text{m} \) deep, is used to pump water uphill to a house that is \( 50 \) \( \text{m} \) above the top of the well, as shown to the right. The density of water is \( 1000 \) \( \text{kg/m}^3 \). All pressures are gauge pressures. Neglect the effects of friction, turbulence, and viscosity.
Two paper cups are suspended by strings and hung near each other. They are separated by about \( 10 \) \( \text{cm} \). Explain what happens to the cups when you blow air between them. Hint: Do they remain still, moves away from each other or move towards each other?
A spherical balloon of mass \( 226 \) \( \text{kg} \) is filled with helium gas until its volume is \( 325 \) \( \text{m}^3 \). Assume the density of air is \( 1.29 \) \( \text{kg/m}^3 \) and the density of helium is \( 0.179 \) \( \text{kg/m}^3 \).
Water flows from point \( A \) to points \( D \) and \( E \) as shown. Some of the flow parameters are known, as shown in the table. Determine the unknown parameters. Note the diagram above does not show the relative diameters of each section of the pipe.
| Section | Diameter | Flow Rate | Velocity |
|---|---|---|---|
| \( \text{AB} \) | \( 300 \) \( \text{mm} \) | \(\textbf{?}\) | \(\textbf{?}\) |
| \( \text{BC} \) | \( 600 \) \( \text{mm} \) | \(\textbf{?}\) | \( 1.2 \) \( \text{m/s} \) |
| \( \text{CD} \) | \(\textbf{?}\) | \( Q_{CD} = 2Q_{CE} \) \( \text{m}^3/\text{s} \) | \( 1.4 \) \( \text{m/s} \) |
| \( \text{CE} \) | \( 150 \) \( \text{mm} \) | \( Q_{CE} = 0.5Q_{CD} \) \( \text{m}^3/\text{s} \) | \(\textbf{?}\) |
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?
Which of the following statements is an expression of the equation of continuity?
Three identical reservoirs, \(A\), \(B\), and \(C\), are represented above, each with a small pipe where water exits horizontally. The pipes are set at the same height above a pool of water. The water in the reservoirs is kept at the levels shown. Which of the following correctly ranks the horizontal distances \( d \) that the streams of water travel before hitting the surface of the pool?
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?
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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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