0 attempts
0% avg
UBQ Credits
Part A – Explanation of forces on FBD
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| a) | \(B\) | This is the buoyant force acting upward on the balloon, equal to the weight of the displaced air. |
| a) | \(W_{\text{balloon}} = (0.0050\,\text{kg})\,g\) | This is the weight of the balloon (including helium) acting downward. |
| a) | \(T\) | This is the tension in the string, which transmits the downward pull from the attached \(0.015\,\text{kg}\) object. (Only the balloon is considered here so its forces include the upward buoyant force and the two downward forces: its own weight and the tension from the string.) |
Part B – Buoyant force on the balloon
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| b) 1 | \(B = W_{\text{balloon}} + W_{\text{object}} = (0.0050\,\text{kg} + 0.015\,\text{kg})\,g = 0.0200\,g\) | The balloon is just prevented from rising, so its upward buoyant force exactly balances the combined weight of the balloon and the attached object. (The buoyant force on the object is negligible.) |
| b) 2 | \(B = 0.0200 \times 9.8 \; (\text{m/s}^2) = \boxed{0.196\,\text{N}}\) | Substitute \(g \approx 9.8\,\text{m/s}^2\) to calculate the buoyant force in newtons. |
Part C – Volume of the balloon
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| c) 1 | \(B = \rho_{\text{air}} \; g \; V\) | This is Archimedes’ principle stating that the buoyant force is equal to the weight of the displaced air. Here \(\rho_{\text{air}} = 1.29\,\text{kg/m}^3\), and \(g = 9.8\,\text{m/s}^2\). |
| c) 2 | \(V = \displaystyle \frac{B}{\rho_{\text{air}} \; g} = \frac{0.196}{1.29 \times 9.8}\) | Rearrange the equation to solve for the volume \(V\) of the balloon. |
| c) 3 | \(V \approx \frac{0.196}{12.642} \approx \boxed{0.0155\,\text{m}^3}\) | Calculate the denominator (\(1.29 \times 9.8 \approx 12.642\)) and then evaluate \(V\). The volume of the balloon is approximately \(0.0155\,\text{m}^3\). |
Part D – Position of ballon in the car
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| d) 1 | Effective gravity: \(\vec{g}_{\text{eff}} = \vec{g} – \vec{a}\) | When the car accelerates forward with acceleration \(a\), the effective gravitational field in the car’s frame tilts backward. This is a standard result for non-inertial frames. |
| d) 2 | For a mass not affected by buoyancy, the equilibrium direction of the string makes an angle \(\theta\) where \(\tan(\theta)=\frac{a}{g}\). | The heavy \(0.015\,\text{kg}\) object behaves as a normal pendulum, aligning along the effective gravitational field. This causes it to deflect opposite to the car’s acceleration (i.e., towards the rear of the car). |
| d) 3 | Result: The \(0.015\,\text{kg}\) object \(\rightarrow\) moves backward relative to the car. | Thus, when the car accelerates forward, the object swings toward the back of the car, while the buoyant balloon (not shown here) would deflect forward. Since the child holds the string midway, the object is pulled into its equilibrium position defined by \(\vec{g}_{\text{eff}}\). |
Just ask: "Help me solve this problem."
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?
Marc’ favorite ride at Busch Gardens 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 \( 30 \)\( \text{cm}^2 \) cylindrical piston, which holds the \( 20,000 \)\( \text{N} \) ride off the ground. What is the area of the piston that holds the ride?
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?
The radius of the left piston is \( 0.12 \) \( \text{m} \) and the radius of the right piston is \( 0.65 \) \( \text{m} \). If \( f \) were raised by \( 14 \) \( \text{N} \), how much would \( F \) need to be increased to maintain equilibrium?
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
