0 attempts
0% avg
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[V_{\text{outer}} = \frac{4}{3}\pi (1.5)^3 \quad , \quad V_{\text{inner}} = \frac{4}{3}\pi r_i^3 \quad , \quad V_{\text{shell}} = \frac{4}{3}\pi\left(1.5^3 – r_i^3\right)\] | Calculate the volume of the full sphere (with outer radius \(1.5\) m) and the hollow part (inner radius \(r_i\)). Their difference gives the volume of titanium used in the shell. |
| 2 | \[\rho_{\text{Ti}}\,\frac{4}{3}\pi\left(1.5^3 – r_i^3\right) = \rho_{\text{water}}\,\frac{4}{3}\pi (1.5)^3\] | For neutral buoyancy, the mass of the titanium shell must equal the mass of the water displaced by the outer sphere. Here, \(\rho_{\text{Ti}}\) is the density of titanium and \(\rho_{\text{water}}\) is the density of water. |
| 3 | \[\rho_{\text{Ti}}\left(1.5^3 – r_i^3\right) = \rho_{\text{water}}\,(1.5)^3\] | Cancel the common factor \(\frac{4}{3}\pi\) from both sides of the equation. |
| 4 | \[1.5^3 – r_i^3 = \frac{\rho_{\text{water}}}{\rho_{\text{Ti}}}(1.5)^3\] | Rearrange the equation to isolate the term \(r_i^3\). |
| 5 | \[r_i^3 = (1.5)^3\left(1 – \frac{\rho_{\text{water}}}{\rho_{\text{Ti}}}\right)\] | Solve for \(r_i^3\) by subtracting and rearranging the terms. |
| 6 | \[r_i^3 = 3.375\left(1 – \frac{1000}{4500}\right) = 3.375\left(\frac{7}{9}\right) \approx 2.625\] | Substitute \(\rho_{\text{Ti}} \approx 4500\,\text{kg/m}^3\) and \(\rho_{\text{water}} \approx 1000\,\text{kg/m}^3\). Note that \((1.5)^3 = 3.375\) and \(1 – \frac{1000}{4500} = \frac{7}{9}\). |
| 7 | \[r_i = \sqrt[3]{2.625} \approx 1.38\,\text{m}\] | Take the cube root of \(r_i^3\) to obtain the inner radius needed for neutral buoyancy. |
| 8 | \[\boxed{r_i \approx 1.38\,\text{m}}\] | This is the final answer for the inner radius of the sphere. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
Find the approximate minimum mass needed for a spherical ball with a \(40\) \(\text{cm}\) radius to sink in a liquid of density \(1.4 \times 10^3\) \(\text{kg/m}^3\). Use \(9.8 \text{m/s}^2\) for \(g\).
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of \( 0.5 \) \( \frac{\text{m}}{\text{s}} \) through a \( 2 \) \( \text{cm} \) diameter pipe in the basement under a pressure of \( 3 \) \( \text{atm} \), what will be the flow speed and pressure in a \( 1.3 \) \( \text{cm} \) diameter pipe on the second floor \( 5 \) \( \text{m} \) above?
In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
A person is standing on a railroad station platform when a high-speed train passes by. The person will tend to be
A solid plastic cube with uniform density (side length = \(0.5\) \(\text{m}\)) of mass \(100\) \(\text{kg}\) is placed in a vat of fluid whose density is \(1200\) \(\text{kg/m}^3\). What fraction of the cube’s volume floats above the surface of the fluid?
A cube of side length \( s \) rests on the bottom surface of a container of fluid. The fluid is at a height \( y \) above the bottom of the tank. The fluid has density \( \rho \) and the atmospheric pressure is \( P_{\text{atm}} \).
Which of the following expressions is equal to the absolute pressure exerted by the fluid on the top surface of the cube?
In the lab, a student is given a glass beaker filled with water with an ice cube of mass \( m \) and volume \( V_c \) floating in it.
The downward force of gravity on the ice cube has magnitude \( F_g \). The student pushes down on the ice cube with a force of magnitude \( F_P \) so that the cube is totally submerged. The water then exerts an upward buoyant force on the ice cube of magnitude \( F_B \). Which of the following is an expression for the magnitude of the acceleration of the ice cube when it is released?
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 diagram above shows a hydraulic chamber with a spring \( (k_s = 1250 \, \text{N/m}) \) attached to the input piston and a rock of mass \( 55.2 \, \text{kg} \) resting on the output plunger. The input piston and output plunger are at about the same height, and each has negligible mass. The chamber is filled with 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 \(1.025 \times 10^3 \text{ kg/m}^3\).
\(r_i \approx 1.38\,\text{m}\)
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
