| 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. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
A small sphere hangs from a string attached to the ceiling of a uniformly accelerating train car. It is observed that the string makes an angle of \(37^\circ\) with respect to the vertical. The magnitude of the acceleration \(a\) of the train car is most nearly:
An air mattress pump blows air above a beach ball at \( 8 \) \( \text{m/s} \). The air below the beach ball is moving at \( \approx 0 \) \( \text{m/s} \). Assuming the beach ball diameter is \( 0.1 \) \( \text{m} \), meaning the areas for the top \& bottom are each \( \approx 0.03 \) \( \text{m}^2 \), and the density of air is \( 1 \) \( \text{kg/m}^3 \), what is the lift force on the beach ball?
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 \( 2.0 \) \( \text{m} \).
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 \( 1500 \) \( \text{kg} \)?
Diamond has a density of \( 3500 \) \( \text{kg/m}^3 \). During a physics lab, a diamond drops out of Virginia’s necklace and falls into her graduated cylinder filled with \( 5.00 \times 10^{-5} \) \( \text{m}^3 \) of water. This causes the water level to rise to the \( 5.05 \times 10^{-5} \) \( \text{m}^3 \) mark. What is the mass of Virginia’s diamond?
Caleb is filling up water balloons for the Physics Olympics balloon toss competition. Caleb sets a \( 0.50 \text{-kg} \) spherical water balloon on the kitchen table and notices that the bottom of the balloon flattens until the pressure on the bottom is reduced to \( 630 \frac{\text{N}}{\text{m}^2} \). What is the area of the flat spot on the bottom of the balloon?
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?
A vertical rope of negligible mass supports a block that weighs \(30 N\). The breaking strength of the rope is \( 50 N\). The largest acceleration that can be given to the block by pulling up on it with the rope without breaking the rope is most nearly
Two blocks of the same size are floating in a container of water. The first block is submerged \( 80\% \) while the second block is submerged by \( 20\% \) beneath the water. Which of the following is a correct statement about the two blocks?
Two points, \( A \) and \( B \), are in a pipe carrying a flowing ideal fluid. Point \( B \) is \( 2.0 \) \( \text{m} \) higher than point \( A \), and the fluid speed at \( B \) is twice the speed at \( A \). If the pressure at \( A \) is \( P_A \), which of the following expressions correctly represents the pressure at \( B \) \( (P_B) \)?
\(1.32\times10^{3}\,\text{N}\)
\(1.02\,\text{m/s}^2\,\text{upward}\)
By continuing you (1) agree to our Terms of Use and Terms of Sale 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] |
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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 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.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
