| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[A_A = \pi\left(\frac{D}{2}\right)^2\] | At Point A, the pipe diameter is \(D\), so the cross-sectional area is found using the area formula for a circle. |
| 2 | \[A_B = \pi\left(\frac{2D}{2}\right)^2 = \pi D^2\] | At Point B, the pipe diameter is \(2D\), so the radius is \(D\), making the area larger than at Point A. |
| 3 | \[A_A = \frac{\pi D^2}{4}\] | Simplifying the area at Point A shows that \(A_B\) is \(4\) times greater than \(A_A\). |
| 4 | \[A_A v_0 = A_B v_B\] | Use the continuity equation for incompressible fluid flow. Since water density is constant, the volume flow rate is the same at Points A and B. |
| 5 | \[\left(\frac{\pi D^2}{4}\right)v_0 = \left(\pi D^2\right)v_B\] | Substitute the cross-sectional areas into the continuity equation. |
| 6 | \[v_B = \frac{v_0}{4}\] | Solve algebraically for the velocity at Point B. Because the area at Point B is \(4\) times larger, the speed at Point B is \(\frac{1}{4}\) of the speed at Point A. |
| 7 | \[P_A + \frac{1}{2}\rho v_A^2 + \rho g y_A = P_B + \frac{1}{2}\rho v_B^2 + \rho g y_B\] | Apply Bernoulli’s equation between Points A and B. The pressure, kinetic energy density, and gravitational potential energy density terms are related along the flow. |
| 8 | \[P_0 + \frac{1}{2}\rho v_0^2 + \rho g y_A = P_B + \frac{1}{2}\rho\left(\frac{v_0}{4}\right)^2 + \rho g(y_A + H)\] | Substitute \(P_A = P_0\), \(v_A = v_0\), \(v_B = \frac{v_0}{4}\), and \(y_B = y_A + H\). |
| 9 | \[P_0 + \frac{1}{2}\rho v_0^2 = P_B + \frac{1}{32}\rho v_0^2 + \rho gH\] | Cancel \(\rho g y_A\) from both sides and simplify \(\frac{1}{2}\rho\left(\frac{v_0}{4}\right)^2 = \frac{1}{32}\rho v_0^2\). |
| 10 | \[P_B = P_0 – \rho gH + \frac{1}{2}\rho v_0^2 – \frac{1}{32}\rho v_0^2\] | Rearrange algebraically to solve for the pressure at Point B. |
| 11 | \[P_B = P_0 – \rho gH + \left(\frac{16}{32} – \frac{1}{32}\right)\rho v_0^2\] | Combine the kinetic energy density terms using a common denominator. |
| 12 | \[\boxed{P_B = P_0 – \rho gH + \frac{15}{32}\rho v_0^2}\] | This matches choice \(\text{(c)}\). Choice \(\text{(a)}\) ignores the speed decrease in the wider pipe. Choice \(\text{(b)}\) uses an incorrect velocity ratio or kinetic term. Choice \(\text{(d)}\) does not follow Bernoulli’s equation and ignores height and flow-speed effects. |
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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 student designs an experiment to determine the density of an unknown fluid. The student pours the fluid into a graduated cylinder and attaches an object to a force probe. The object has a density greater than the density of the fluid. The student partially submerges the object into the fluid and records both the volume of fluid displaced in the graduated cylinder and the reading on the force probe. The student then submerges the object further and, at each trial, records the new values of displaced volume and force probe reading until the object is fully submerged. The student constructs a graph of force probe reading (vertical axis) as a function of volume of fluid displaced (horizontal axis). Which of the following statements correctly describes how a feature of this graph is related to the density of the fluid?
Why do you float higher in salt water than in fresh water?
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?
Diamond has a density of \( 3500 \) \( \text{kg/m}^3 \). During a physics lab, a diamond drops out of Virginia’s necklace and falls into her graduated cylinder filled with \( 5.00 \times 10^{-5} \) \( \text{m}^3 \) of water. This causes the water level to rise to the \( 5.05 \times 10^{-5} \) \( \text{m}^3 \) mark. What is the mass of Virginia’s diamond?
How large must a heating duct be if air moving \( 3 \ \frac{\text{m}}{\text{s}} \) along it can replenish the air in a room of \( 300 \ \text{m}^3 \) volume every \( 15 \) minutes? Assume the air’s density remains constant.
An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
The difference in pressure between the atmosphere and the human lungs is \( 1.05 \times 10^5 \) \( \text{Pa} \). What is the longest straw you could use to draw up milk whose density is \( 1030 \) \( \text{kg/m}^3 \)?
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?
The figure shows a horizontal pipe with sections with different cross-sectional areas. Small tubes extend from the top of each section. The cross-sectional area of the pipe at location C is half that at A, and the areas at A and D are the same. Water flows in the pipe from left to right. Which of the following correctly ranks the height \( h \) of the water in the tubes above the labeled locations?
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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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