| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[m_{He}=\rho_{He}V\] | Mass of helium found from \(m=\rho V\). |
| 2 | \[m_{He}=0.179\,(\text{kg/m}^3)(325\,\text{m}^3)=58.2\,\text{kg}\] | Substitute the given values. |
| 3 | \[m_{tot}=m_{bal}+m_{He}\] | Total mass equals envelope plus helium. |
| 4 | \[m_{tot}=226\,\text{kg}+58.2\,\text{kg}=284.2\,\text{kg}\] | Add the masses. |
| 5 | \[W=m_{tot}g\] | Weight of balloon system, where \(g=9.8\,\text{m/s}^2\). |
| 6 | \[W=(284.2\,\text{kg})(9.8\,\text{m/s}^2)=2.78\times10^{3}\,\text{N}\] | Compute the weight. |
| 7 | \[F_B=\rho_{air}Vg\] | Buoyant force equals the weight of displaced air. |
| 8 | \[F_B=(1.29\,\text{kg/m}^3)(325\,\text{m}^3)(9.8\,\text{m/s}^2)=4.11\times10^{3}\,\text{N}\] | Insert values for air density and volume. |
| 9 | \[T=F_B-W\] | With equilibrium (no motion) and upward positive: \(F_B-W-T=0\). |
| 10 | \[T=4.11\times10^{3}\,\text{N}-2.78\times10^{3}\,\text{N}=1.32\times10^{3}\,\text{N}\] | Calculate the tension magnitude. |
| 11 | \[\boxed{T\approx1.32\times10^{3}\,\text{N}\;\text{(downward)}}\] | The string pulls downward with this tension. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[m_{new}=m_{bal}+m_{He}+m_{bas}\] | Total mass now includes basket mass \(m_{bas}=95.5\,\text{kg}\). |
| 2 | \[m_{new}=226+58.2+95.5=379.7\,\text{kg}\] | Add the three masses. |
| 3 | \[W_{new}=m_{new}g\] | Compute new total weight. |
| 4 | \[W_{new}=(379.7\,\text{kg})(9.8\,\text{m/s}^2)=3.72\times10^{3}\,\text{N}\] | Numerical value of weight. |
| 5 | \[F_{net}=F_B-W_{new}\] | Upward net force after string is cut. |
| 6 | \[F_{net}=4.11\times10^{3}\,\text{N}-3.72\times10^{3}\,\text{N}=3.88\times10^{2}\,\text{N}\] | Subtract the forces. |
| 7 | \[a=\frac{F_{net}}{m_{new}}\] | Newton’s second law \(\sum F = m a\). |
| 8 | \[a=\frac{3.88\times10^{2}\,\text{N}}{379.7\,\text{kg}}=1.02\,\text{m/s}^2\] | Divide to find acceleration. |
| 9 | \[\boxed{a\approx1.02\,\text{m/s}^2\;\text{upward}}\] | Magnitude and upward direction of acceleration. |
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A conical pendulum is formed by attaching a ball of mass \( m \) to a string of length \( \ell \), then allowing the ball to move in a horizontal circle of radius \( r \). The following figure shows that the string traces out the surface of a cone, hence the name.
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?
A geologist suspects that her rock specimen is hollow, so she weighs the specimen in both air and water. When completely submerged, the rock weighs twice as much in air as it does in water.
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, \( \dfrac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
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?
The \( 70 \) \( \text{kg} \) student in the figure balances a \( 1200 \) \( \text{kg} \) elephant on a hydraulic lift. Assume that it is filled with oil, which is incompressible and has a density \( \rho = 900 \) \( \text{kg/m}^3 \). What is the diameter of the piston the student is standing on? Assume each piston has a cylindrical shape, i.e., a circular cross-sectional area. Note: The two pistons are at the same height. Also, the diameter of the wider piston is given in the figure to be \( 2.0 \) \( \text{m} \).
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.
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 \(g\).
A fountain with an opening of radius \( 0.015 \) \( \text{m} \) shoots a stream of water vertically from ground level at \( 6.0 \) \( \text{m/s} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
\(1.32\times10^{3}\,\text{N}\)
\(1.02\,\text{m/s}^2\,\text{upward}\)
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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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