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| Derivation / Formula | Reasoning |
|---|---|
| \[P_g = \rho g h\] | Hydrostatic gauge pressure at depth \(h\) in a fluid of density \(\rho\). |
| \[P_g = (1.025\times10^{3})\,(9.8)\,(35)\] | Substitute \(\rho = 1.025\times10^{3}\;\text{kg/m}^3\), \(g = 9.8\;\text{m/s}^2\), \(h = 35\;\text{m}\). |
| \[\boxed{P_g \approx 3.5\times10^{5}\;\text{Pa}}\] | Numeric evaluation gives the gauge pressure. |
| Derivation / Formula | Reasoning |
|---|---|
| \[P_{abs} = P_{atm} + P_g\] | Absolute pressure equals atmospheric plus gauge pressure. |
| \[P_{abs} = 1.01\times10^{5} + 3.5\times10^{5}\] | Add standard atmospheric pressure \(P_{atm}=1.01\times10^{5}\;\text{Pa}\). |
| \[\boxed{P_{abs} \approx 4.5\times10^{5}\;\text{Pa}}\] | Numeric evaluation of absolute pressure. |
| Derivation / Formula | Reasoning |
|---|---|
| \[V = 1.0\times2.0\times0.03 = 0.06\;\text{m}^3\] | Volume of the rectangular plate. |
| \[m = \rho_{Al} V = (2.7\times10^{3})(0.06)\] | Mass from density of aluminum \(\rho_{Al}\). |
| \[m = 162\;\text{kg}\] | Numeric result for mass. |
| \[W = mg = 162\,(9.8)\] | Weight of the plate. |
| \[W = 1.59\times10^{3}\;\text{N}\] | Numeric value of weight. |
| \[F_b = \rho_{w} g V = (1.025\times10^{3})(9.8)(0.06)\] | Buoyant force using Archimedes’ principle, \(\rho_{w}\) is water density. |
| \[F_b \approx 6.0\times10^{2}\;\text{N}\] | Numeric value of the buoyant force. |
| \[T = W – F_b\] | For slow constant upward motion, net force is zero, so tension balances weight minus buoyancy. |
| \[\boxed{T \approx 9.8\times10^{2}\;\text{N}}\] | Calculated cable tension. |
| Derivation / Formula | Reasoning |
|---|---|
| \[T’ = W – F_b + m a\] | Newton’s second law: upward acceleration \(a\) requires extra upward force \(m a\). |
| \[m a > 0 \;\Rightarrow\; T’ > T\] | Since \(a = 0.05\;\text{m/s}^2\) is upward, the added term increases tension. |
| \[\boxed{T’ \text{ increases}}\] | Therefore, the cable tension becomes larger when the plate accelerates upward. |
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Water flows from point \( A \) to points \( D \) and \( E \) as shown. Some of the flow parameters are known, as shown in the table. Determine the unknown parameters. Note the diagram above does not show the relative diameters of each section of the pipe.
| Section | Diameter | Flow Rate | Velocity |
|---|---|---|---|
| \( \text{AB} \) | \( 300 \) \( \text{mm} \) | \(\textbf{?}\) | \(\textbf{?}\) |
| \( \text{BC} \) | \( 600 \) \( \text{mm} \) | \(\textbf{?}\) | \( 1.2 \) \( \text{m/s} \) |
| \( \text{CD} \) | \(\textbf{?}\) | \( Q_{CD} = 2Q_{CE} \) \( \text{m}^3/\text{s} \) | \( 1.4 \) \( \text{m/s} \) |
| \( \text{CE} \) | \( 150 \) \( \text{mm} \) | \( Q_{CE} = 0.5Q_{CD} \) \( \text{m}^3/\text{s} \) | \(\textbf{?}\) |
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?
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.
Marc’s favorite ride at Busch Gardens is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a force of \( 72 \) \( \text{N} \) to a \( 30 \)\( \text{cm}^2 \) cylindrical piston, which holds the \( 20,000 \)\( \text{N} \) ride off the ground. What is the area of the piston that holds the ride?
Why do you float higher in salt water than in fresh water?
Ben’s favorite ride at the Barrel-O-Fun Amusement Park is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a force of \( 72 \) \( \text{N} \) to a \( 3.0 \) \( \text{cm} \) wide cylindrical piston, which holds the \( 20,000 \) \( \text{N} \) ride off the ground. What is the diameter of the piston that holds the ride?
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 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.
Water flowing in a horizontal pipe speeds up as it goes from a section with a large diameter to a section with a small diameter. Which of the following can explain why the speed of the water increases?
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 \(3\) \(\text{cm}\).
\(P_g \approx 3.5\times10^{5}\,\text{Pa}\)
\(P_{abs} \approx 4.5\times10^{5}\,\text{Pa}\)
\(T \approx 9.8\times10^{2}\,\text{N}\)
\(T \text{ increases}\)
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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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