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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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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?
A solid titanium sphere of radius \( 0.35 \) \( \text{m} \) has a density \( 4500 \) \( \text{kg/m}^3 \). It is held suspended completely underwater by a cable. What is the tension in the cable?
A crane’s trolley at point \( P \) moves for a few seconds to the right with constant acceleration, and the \( 870 \, \text{kg} \) load hangs on a light cable at a \( 5^\circ \) angle to the vertical as shown. What is the acceleration of the trolley and load?
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.
Ben’s favorite ride at the Barrel-O-Fun Amusement Park is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a force of \( 72 \) \( \text{N} \) to a \( 3.0 \) \( \text{cm} \) wide cylindrical piston, which holds the \( 20,000 \) \( \text{N} \) ride off the ground. What is the diameter of the piston that holds the ride?
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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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