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| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | [\displaystyle F_{B}=\rho_{\text{air}}\,V\,g] | Compute the buoyant force where \(\rho_{\text{air}}=1.29\,\text{kg/m}^3\), \(V=325\,\text{m}^3\), and \(g=9.8\,\text{m/s}^2\). |
| 2 | [\displaystyle F_{B}=1.29\times325\times9.8] | Substitute the given values to evaluate the buoyant force. |
| 3 | [\displaystyle F_{B}\approx4108.65\,\text{N}] | Resulting upward force due to buoyancy. |
| 4 | [\displaystyle m_{He}=\rho_{He}\,V] | Calculate the mass of helium inside the balloon using \(\rho_{He}=0.179\,\text{kg/m}^3\) and \(V=325\,\text{m}^3\). |
| 5 | [\displaystyle m_{He}=0.179\times325\approx58.175\,\text{kg}] | Evaluate the helium mass. |
| 6 | [\displaystyle m_{\text{total}}=226+58.175] | Add the balloon envelope mass (226 kg) and the helium mass to get the total mass. |
| 7 | [\displaystyle m_{\text{total}}\approx284.175\,\text{kg}] | Computed total mass of the balloon system. |
| 8 | [\displaystyle W=m_{\text{total}}\,g] | Calculate the weight of the balloon system. |
| 9 | [\displaystyle W\approx284.175\times9.8\approx2783.515\,\text{N}] | Evaluate the downward gravitational force. |
| 10 | [\displaystyle T=F_{B}-W] | For equilibrium, the tension in the string balances the net upward force. |
| 11 | [\displaystyle T\approx4108.65-2783.515\approx1325.135\,\text{N}] | Calculate the magnitude of the tension in the string. |
| 12 | [\displaystyle \boxed{T\approx1325\,\text{N}}] | Final answer for part (a); the string must exert approximately 1325 N downward. |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | [\displaystyle m_{\text{basket}}=95.5\,\text{kg}] | Identify the mass of the basket added to the system. |
| 2 | [\displaystyle m_{\text{new}}=m_{\text{total}}+m_{\text{basket}}] | Add the basket mass to the previous total mass of the balloon system. |
| 3 | [\displaystyle m_{\text{new}}\approx284.175+95.5\approx379.675\,\text{kg}] | Computed overall mass of the balloon with the basket. |
| 4 | [\displaystyle W_{\text{new}}=m_{\text{new}}\,g] | Calculate the total weight of the new system. |
| 5 | [\displaystyle W_{\text{new}}\approx379.675\times9.8\approx3719.415\,\text{N}] | Determine the downward gravitational force for the combined mass. |
| 6 | [\displaystyle F_{\text{net}}=F_{B}-W_{\text{new}}] | Find the net upward force acting on the system; note that \(F_{B}\) remains unchanged. |
| 7 | [\displaystyle F_{\text{net}}\approx4108.65-3719.415\approx389.235\,\text{N}] | Evaluate the net force after adding the basket. |
| 8 | [\displaystyle a=\frac{F_{\text{net}}}{m_{\text{new}}}] | Determine the acceleration using Newton’s second law \(F=ma\). |
| 9 | [\displaystyle a\approx\frac{389.235}{379.675}\approx1.0248\,\text{m/s}^2] | Calculate the magnitude of the vertical acceleration. |
| 10 | [\displaystyle \boxed{a\approx1.02\,\text{m/s}^2\text{ upward}}] | Final answer for part (b); the acceleration is approximately 1.02 m/s² upward. |
Just ask: "Help me solve this problem."
You have a giant cask of water with a spigot some height below the water surface. The surface of the water, which is essentially at rest, is exposed to atmosphere (\( \approx 10^5 \text{Pa} \)). The water density is \( \approx 1000 \text{kg/m}^3 \). The water pours out of the spigot at \( 3 \text{m/s} \). How far below the water surface is the spigot positioned?
A fluid flows through the two sections of cylindrical pipe shown in the figure. The narrow section of the pipe has radius \( R \) and the wide section has radius \( 2R \). What is the ratio of the fluid’s speed in the wide section of pipe to its speed in the narrow section of pipe, \( \frac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
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 pair of fuzzy dice is hanging by a string from your rearview mirror. You speed up from a stoplight. During the acceleration, the dice do not move vertically; the string makes an angle of \( 22^\circ \) with the vertical. The dice have a mass of \( 0.10 \, \text{kg} \). Determine the acceleration.
A trash compactor pushes down with a force of \( 500 \) \( \text{N} \) on a \( 3 \) \( \text{cm}^2 \) input piston, causing a force of \( 30,000 \) \( \text{N} \) to crush the trash. What is the area of the output piston that crushes the trash?
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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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