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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. |
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Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
Two wires support an unknown mass as shown in the diagram. The tension in the left wire is measured to be \( 17.5 \) \( \text{N} \) and the tension in the right wire is \( 30.3 \) \( \text{N} \). The left wire makes an angle of \( 30^{\circ} \) with the vertical, and the right wire makes an angle of \( 60^{\circ} \) with the vertical. What is the mass of the object?
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?
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.
Two wires are tied to the \(500 \, \text{g}\) sphere as shown above. The sphere revolves in a horizontal circle at a constant speed of \(7.2 \, \text{m/s}\). What is the tension in the upper wire? What is the tension in the lower wire?
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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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