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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 beaker weighing \( 2.0 \) \( \text{N} \) is filled with \( 5.0 \times 10^{-3} \) \( \text{m}^3 \) of water. A rubber ball weighing \( 3.0 \) \( \text{N} \) is held entirely underwater by a massless string attached to the bottom of the beaker, as represented in the figure above. The tension in the string is \( 4.0 \) \( \text{N} \). The water fills the beaker to a depth of \( 0.20 \) \( \text{m} \). Water has a density of \( 1000 \) \( \text{kg/m}^3 \). The effects of atmospheric pressure may be neglected.
A drinking fountain projects water at an initial angle of \( 50^ \circ \) above the horizontal, and the water reaches a maximum height of \( 0.150 \) \( \text{m} \) above the point of exit. Assume air resistance is negligible.
Two objects labeled K and L have equal mass but densities \( 0.95D_o \) and \( D_o \), respectively. Each of these objects floats after being thrown into a deep swimming pool. Which is true about the buoyant forces acting on these objects?
A spherical balloon has a radius of \(7.15\) \(\text{m}\) and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of \(930\) \(\text{kg}\)?
Take the density of helium and air to be \(0.18\) \(\text{kg/m}^3\) and \(1.24\) \(\text{kg/m}^3\), respectively.
A sphere of mass \(0.5\) \(\text{kg}\) is dropped into a column of oil. At the instant the sphere becomes completely submerged in the oil, the sphere is moving downward at \(8\) \(\text{m/s}\), the buoyancy force on the sphere is \(4.0\) \(\text{N}\), and the fluid frictional force is \(4.0\) \(\text{N}\). Which of the following describes the motion of the sphere at this instant?
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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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