0 attempts
0% avg
UBQ Credits
| Derivation or Formula | Reasoning |
|---|---|
| \[A_1 v_i = A_2 v_x\] | Conservation of mass (continuity equation) requires that the volumetric flow rate remains constant, where \(A_1\) and \(A_2\) are the cross-sectional areas of the basement and second floor pipes, respectively. |
| \[A = \frac{\pi}{4}d^2\] | This is the formula for the area of a circle, with \(d\) as the diameter of the pipe. We use it to express \(A_1\) and \(A_2\) in terms of the given diameters. |
| \[v_x = \left(\frac{A_1}{A_2}\right)v_i = \left(\frac{d_1}{d_2}\right)^2 v_i\] | Substituting the area formula into the continuity equation, the constant factors cancel and we obtain \(v_x\) in terms of the diameters \(d_1\) and \(d_2\) and the initial speed \(v_i\). |
| \[v_x = \left(\frac{0.02}{0.013}\right)^2 (0.5)\] | Substitute \(d_1 = 0.02\,\text{m}\), \(d_2 = 0.013\,\text{m}\), and \(v_i = 0.5\,\text{m/s}\) into the equation. |
| \[v_x \approx (1.5385)^2 \times 0.5 \approx 2.367 \times 0.5 \approx 1.18\,\text{m/s}\] | Evaluating the ratio gives approximately \(1.5385\) which squared is about \(2.367\). Multiplying by \(0.5\,\text{m/s}\) results in a speed of approximately \(1.18\,\text{m/s}\) in the second floor pipe. |
| \[\boxed{v_x \approx 1.18\,\text{m/s}}\] | This is the final expression for the flow speed in the 1.3 cm diameter pipe on the second floor. |
| Derivation or Formula | Reasoning |
|---|---|
| \[P_1 + \frac{1}{2}\rho v_i^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_x^2 + \rho g h_2\] | This is Bernoulli’s equation applied between the basement (point 1) and the second floor (point 2). It relates pressure, kinetic, and potential energy per unit volume along a streamline. |
| \[h_1 = 0,\quad h_2 = 5\,\text{m}\] | We set the basement as the reference level (\(h_1 = 0\)) so that the second floor is at an elevation of \(5\,\text{m}\). |
| \[P_1 = 3\,\text{atm} = 3 \times 101325 = 303975\,\text{Pa}\] | The initial pressure is given as \(3\,\text{atm}\); converting to Pascals using \(1\,\text{atm} = 101325\,\text{Pa}\) yields \(303975\,\text{Pa}\). |
| \[P_2 = P_1 + \frac{1}{2}\rho (v_i^2 – v_x^2) – \rho g (h_2 – h_1)\] | Rearrange Bernoulli’s equation to solve for \(P_2\), the pressure in the second floor pipe. |
| \[P_2 = 303975 + \frac{1}{2}(1000)(0.5^2 – 1.18^2) – (1000)(9.8)(5)\] | Substitute \(\rho = 1000\,\text{kg/m}^3\) for water, \(g = 9.8\,\text{m/s}^2\), \(v_i = 0.5\,\text{m/s}\), and \(v_x \approx 1.18\,\text{m/s}\) into the equation. |
| \[0.5^2 = 0.25,\quad 1.18^2 \approx 1.392\] | Calculate the squares of the velocities for the kinetic energy terms. |
| \[\frac{1}{2}\times 1000\times(0.25 – 1.392) \approx 500\times(-1.142) \approx -571\,\text{Pa}\] | The kinetic energy difference contributes a change of approximately \(-571\,\text{Pa}\) (negative since the speed increases in the smaller pipe). |
| \[1000 \times 9.8 \times 5 = 49000\,\text{Pa}\] | This term accounts for the increase in gravitational potential energy due to the 5 m elevation gain. |
| \[P_2 = 303975 – 571 – 49000 \approx 254404\,\text{Pa}\] | Subtract the kinetic and potential energy contributions from the initial pressure to find the pressure at the second floor. |
| \[\boxed{P_2 \approx 254400\,\text{Pa}}\] | This is the final pressure in the 1.3 cm diameter pipe on the second floor. In atmospheres, this is roughly \(254400/101325 \approx 2.51\,\text{atm}\). |
Just ask: "Help me solve this problem."
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 \( 1000 \) \( \text{kg/m}^3 \).
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?
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?
A helium-filled balloon is attached by a string of negligible mass to a small \(0.015 \ \text{kg}\) object that is just heavy enough to keep the balloon from rising. The total mass of the balloon, including the helium, is \(0.0050 \ \text{kg}\). The density of air is \(\rho_{\text{air}} = 1.29 \ \text{kg/m}^3\), and the density of helium is \(\rho_{\text{He}} = 0.179 \ \text{kg/m}^3\). The buoyant force on the \(0.015 \ \text{kg}\) object is small enough to be negligible.
A fluid flows through the two sections of cylindrical pipe shown in the figure. The narrow section of the pipe has radius \( R \) and the wide section has radius \( 2R \). What is the ratio of the fluid’s speed in the wide section of pipe to its speed in the narrow section of pipe, \( \frac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
\(1.18\,\text{m/s}\)\n\(254400\,\text{Pa}\)
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] |
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
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
