| 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 \(254400/101325 \approx 2.51\,\text{atm}\). |
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In the laboratory, you are given a cylindrical beaker containing a fluid and you are asked to determine the density \( \rho \) of the fluid. You are to use a spring of negligible mass and unknown spring constant \( k \) that is attached to a vertical stand.
How large must a heating duct be if air moving \( 3 \ \frac{\text{m}}{\text{s}} \) along it can replenish the air in a room of \( 300 \ \text{m}^3 \) volume every \( 15 \) minutes? Assume the air’s density remains constant.
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?
A liquid flows at a constant flow rate through a pipe with circular cross-sections of varying diameters. At one point in the pipe, the diameter is \(2\) \(\text{cm}\) and the flow speed is \(18\) \(\text{m/s}\). What is the flow speed at another point in this pipe, where the diameter is \(3\) \(\text{cm}\).
Wanda watches the fish in her fish tank and notices that the angelfish like to feed at the water’s surface, while the catfish feed \( 0.300 \) \( \text{m} \) below at the bottom of the tank. If the average density of the water in the tank is \( 1000\) \( \text{kg/m}^3 \), what is the absolute pressure on the catfish?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
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, \( \frac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
The difference in pressure between the atmosphere and the human lungs is \( 1.05 \times 10^5 \) \( \text{Pa} \). What is the longest straw you could use to draw up milk whose density is \( 1030 \) \( \text{kg/m}^3 \)?
The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is \( 2.5 \) \( \text{m} \), and the puncture is \( 1 \) \( \text{m} \) above the ground level, what is the efflux speed of the water?
\(1.18\,\text{m/s}\)\n\(254400\,\text{Pa}\)
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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] |
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] |
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