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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."
In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
Two objects labeled K and L have equal mass but densities \( 0.95D_o \) and \( D_o \), respectively. Each of these objects floats after being thrown into a deep swimming pool. Which is true about the buoyant forces acting on these objects?
A solid plastic cube with uniform density (side length = \(0.5\) \(\text{m}\)) of mass \(100\) \(\text{kg}\) is placed in a vat of fluid whose density is \(1200\) \(\text{kg/m}^3\). What fraction of the cube’s volume floats above the surface of the fluid?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
The cart with mass \( M = 3 \, \text{kg} \) is pulled by a massless string and moving on a horizontal track. A weight with mass \( m = 1 \, \text{kg} \) is hung from the other end of the string through a pulley system. Due to the gravitational force acting on the weight of mass \( m \), the cart is accelerated to the left. Find the tension in the string.
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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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