| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(V = \frac{4}{3}\pi r^3\) | Calculate the volume of the balloon. The formula for the volume of a sphere is \(\frac{4}{3}\pi r^3\) where \(r\) is the radius. |
| 2 | \(V = \frac{4}{3}\pi (7.15)^3 \approx 1532.16 \, \text{m}^3\) | Substitute the given radius \(r = 7.15 \, \text{m}\) into the formula to find the volume \(V\). |
| 3 | \(\text{Buoyant force} = \rho_{\text{air}} \cdot V \cdot g\) | Calculate the buoyant force using the formula: buoyant force equals the product of the density of air, volume, and gravitational acceleration \(g = 9.81 \, \text{m/s}^2\). |
| 4 | \(\text{Buoyant force} = 1.24 \cdot 1532.16 \cdot 9.81 \approx 18632.06 \, \text{N}\) | Substitute the values: \(\rho_{\text{air}} = 1.24 \, \text{kg/m}^3\), \(V = 1532.16 \, \text{m}^3\), and \(g = 9.81 \, \text{m/s}^2\) into the buoyant force equation. |
| 5 | \(\text{Weight of helium} = \rho_{\text{He}} \cdot V \cdot g\) | Calculate the weight of the helium inside the balloon using the formula: weight equals the product of the density of helium, volume of the balloon, and gravitational acceleration. |
| 6 | \(\text{Weight of helium} = 0.18 \cdot 1532.16 \cdot 9.81 \approx 2706.65 \, \text{N}\) | Substitute the values: \(\rho_{\text{He}} = 0.18 \, \text{kg/m}^3\), \(V = 1532.16 \, \text{m}^3\), and \(g = 9.81 \, \text{m/s}^2\) into the weight of helium equation. |
| 7 | \(\text{Net lift} = \text{Buoyant force} – \text{Weight of helium} – \text{Weight of balloon skin and structure}\) | Calculate the net lifting force by subtracting the weight of the helium and the weight of the balloon skin and structure from the buoyant force. |
| 8 | \(\text{Weight of balloon} = 930 \cdot 9.81 \approx 9123.3 \, \text{N}\) | Calculate the weight of the balloon skin and structure using its mass and gravitational acceleration. |
| 9 | \(\text{Net lift} = 18632.06 – 2706.65 – 9123.3 \approx 6802.11 \, \text{N}\) | Subtract the weight of the helium and the weight of the balloon from the buoyant force to find the net lift force. |
| 10 | \(\text{Maximum cargo mass} = \frac{\text{Net lift}}{g}\) | Calculate the total cargo mass the balloon can lift using the net lift force and dividing by gravitational acceleration \(g\). |
| 11 | \(\text{Maximum cargo mass} = \frac{6802.11}{9.81} \approx 693.23 \, \text{kg}\) | Compute the cargo mass that the balloon is capable of lifting. |
| 12 | \(693.23 \, \text{kg}\) | Therefore, the maximum cargo mass the balloon can lift is \(\boxed{693.23 \, \text{kg}}\). |
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An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
A person is standing on a railroad station platform when a high-speed train passes by. The person will tend to be
Why do you float higher in salt water than in fresh water?
A diver descends from a salvage ship to the ocean floor at a depth of \(35 \text{ m}\) below the surface. The density of ocean water is \(1.025 \times 10^3 \text{ kg/m}^3\).
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?
The large piston in a hydraulic lift has a radius of \( 250 \) \( \text{cm}^2 \). What force must be applied to the small piston with a radius of \( 25 \) \( \text{cm}^2 \) in order to raise a car of mass \( 1500 \) \( \text{kg} \)?
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.
When the button of a trash compactor is pushed, a force of \( 350 \) \( \text{N} \) pushes down on a \( 1.3 \) \( \text{cm}^2 \) input piston, creating a force of \( 22,076 \) \( \text{N} \) to crush the trash. What is the area of the piston that crushes the trash?
The difference in pressure between the atmosphere and the human lungs is \( 1.05 \times 10^5 \) \( \text{Pa} \). What is the longest straw you could use to draw up milk whose density is \( 1030 \) \( \text{kg/m}^3 \)?
An ideal fluid flows through a pipe with radius \( Q \) and flow speed \( V \). If the pipe splits up into three separate paths, each with radius \( \frac{Q}{2} \), what is the flow speed through each of the paths?
\(\boxed{693.23 \, \text{kg}}\)
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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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