| 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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Alcohol has a specific gravity of \( 0.79 \). If a barometer consisting of an open-ended tube placed in a dish of alcohol is used at sea level, to what height in the tube will the alcohol rise?
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?
The drawing above shows a spherical reservoir that contains \( 455,000 \) \( \text{kg} \) of water when full. The reservoir is vented to the atmosphere at the top. Assuming the reservoir is full and the diameter of the reservoir is much larger than any of the pipes on the ground.
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?
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 \(2\)-N force is used to push a small piston \(10\) \(\text{cm}\) downward in a simple hydraulic machine. If the opposite large piston rises by \(0.5\) \(\text{cm}\), what is the maximum weight the large piston can lift?
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 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.
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
A geologist suspects that her rock specimen is hollow, so she weighs the specimen in both air and water. When completely submerged, the rock weighs twice as much in air as it does in 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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