| Derivation or Formula | Reasoning |
|---|---|
| \[F = m \times g = 0.50 \times 9.8 = 4.9\ \text{N}\] | Calculate the weight (force) of the balloon using \(m = 0.50\ \text{kg}\) and gravitational acceleration \(g = 9.8\ \text{m/s}^2\). |
| \[P = \frac{F}{A}\] | This is the definition of pressure, where \(P\) is pressure, \(F\) is force, and \(A\) is area. |
| \[A = \frac{F}{P} = \frac{4.9}{630}\] | Rearrange the pressure formula to solve for the area \(A\), substituting the known force \(4.9\ \text{N}\) and the given pressure \(630\ \text{N/m}^2\). |
| \[A \approx 0.00778\ \text{m}^2\] | Perform the division to find the area of the flat spot on the bottom of the balloon. |
| \[\boxed{0.00778\ \text{m}^2}\] | This is the final answer: the area of the flat spot. |
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An ideal fluid flows through a pipe with radius \( Q \) and flow speed \( V \). If the pipe splits up into three separate paths, each with radius \( \frac{Q}{2} \), what is the flow speed through each of the paths?
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 \(1025\) \(\text{kg/m}^3\).
In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
Water of density \( \rho \) flows through the section of circular pipe shown in the figure. At Point A, where the diameter of the pipe is \( D \), the water has a pressure \( P_0 \) and velocity \( v_0 \). Point B is located a vertical distance \( H \) above Point A in a section of the pipe that has diameter \( 2D \).
Which of the following expressions is equal to the pressure of the water at Point B?
A cube of unknown material and uniform density floats in a container of water with \(60\%\) of its volume submerged. If this same cube were placed in a container of oil with density \(800\) \(\text{kg/m}^3\), what portion of the cube’s volume would be submerged while floating?
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?
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 \(g\).
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 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.
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 \(1.025 \times 10^3 \text{ kg/m}^3\).
\(\boxed{0.00778\,\text{m}^2}\)
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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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