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| 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. |
Just ask: "Help me solve this problem."
Suppose we wish to make a neutrally buoyant hollow sphere out of titanium (\(\rho = 4500 \text{kg/m}^3\)). If the sphere has an outer radius of \( 1.5 \) \( \text{m} \), what must be its inner radius?
A person pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?
Alcohol has a specific gravity of \( 0.79 \). If a barometer consisting of an open-ended tube placed in a dish of alcohol is used at sea level, to what height in the tube will the alcohol rise?
Three identical reservoirs, \(A\), \(B\), and \(C\), are represented above, each with a small pipe where water exits horizontally. The pipes are set at the same height above a pool of water. The water in the reservoirs is kept at the levels shown. Which of the following correctly ranks the horizontal distances \( d \) that the streams of water travel before hitting the surface of the pool?
A block of weight \( W \) is pulled along a horizontal surface at constant speed by a force \( F \), which acts at an angle of \( \theta \) with the horizontal. The normal force exerted on the block by the surface has magnitude:
\(1.32\times10^{3}\,\text{N}\)
\(1.02\,\text{m/s}^2\,\text{upward}\)
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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] |
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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