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Part A
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[P_{\text{gauge}} = \rho\,g\,\Delta x\] | Gauge pressure at a depth is calculated as the product of the fluid’s density \(\rho\), gravitational acceleration \(g\), and the depth \(\Delta x\). |
| 2 | \[P_{\text{gauge}} = (1.025 \times 10^3\,\text{kg/m}^3)(9.8\,\text{m/s}^2)(35\,\text{m})\] | Substitute the given density of ocean water, gravitational acceleration, and depth into the equation. |
| 3 | \[P_{\text{gauge}} \approx 3.53 \times 10^5\,\text{Pa}\] | Performing the multiplication gives the gauge pressure on the ocean floor as approximately \(3.53 \times 10^5\,\text{Pa}\). |
Part B
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[P_{\text{absolute}} = P_{\text{atm}} + P_{\text{gauge}}\] | Absolute pressure is the sum of the atmospheric pressure and the gauge pressure. |
| 2 | \[P_{\text{absolute}} = (1.01 \times 10^5\,\text{Pa}) + (3.53 \times 10^5\,\text{Pa})\] | Substitute the atmospheric pressure and the previously calculated gauge pressure into the formula. |
| 3 | \[P_{\text{absolute}} \approx 4.54 \times 10^5\,\text{Pa}\] | The sum gives the absolute pressure on the ocean floor, approximately \(4.54 \times 10^5\,\text{Pa}\). |
Part C
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[V = 1.0\,\text{m} \times 2.0\,\text{m} \times 0.03\,\text{m} = 0.06\,\text{m}^3\] | Calculate the volume \(V\) of the rectangular aluminum plate using its given dimensions. |
| 2 | \[m_{\text{Al}} = \rho_{\text{Al}}\,V = (2.7 \times 10^3\,\text{kg/m}^3)(0.06\,\text{m}^3) \approx 162\,\text{kg}\] | Determine the mass of the plate by multiplying its volume with the density of aluminum \(\rho_{\text{Al}}\). |
| 3 | \[W = m_{\text{Al}}\,g = 162\,\text{kg} \times 9.8\,\text{m/s}^2 \approx 1588\,\text{N}\] | Compute the weight \(W\) of the plate using \(W = m \times g\). |
| 4 | \[F_B = \rho_{\text{water}}\,g\,V = (1.025 \times 10^3\,\text{kg/m}^3)(9.8\,\text{m/s}^2)(0.06\,\text{m}^3) \approx 603\,\text{N}\] | Calculate the buoyant force \(F_B\) acting on the plate using Archimedes’ principle. |
| 5 | \[T = W – F_B = 1588\,\text{N} – 603\,\text{N} \approx 985\,\text{N}\] | Since the plate is lifted at a constant velocity, the tension \(T\) in the cable equals the net downward force \(W – F_B\). |
| 6 | \[\boxed{T \approx 985\,\text{N}}\] | This is the final tension in the cable required to lift the plate upward at a slow constant velocity. |
Part D
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T’ = (W – F_B) + m_{\text{Al}}\,a\] | When accelerating upward, the cable must supply additional force \(m_{\text{Al}}a\) to overcome inertia, added to the weight minus buoyant force. |
| 2 | \[m_{\text{Al}}\,a = 162\,\text{kg} \times 0.05\,\text{m/s}^2 = 8.1\,\text{N}\] | Calculate the extra force required for an upward acceleration of \(0.05\,\text{m/s}^2\). |
| 3 | \[T’ = 985\,\text{N} + 8.1\,\text{N} \approx 993\,\text{N}\] | Add the extra force to the original tension to find the new tension in the cable. |
| 4 | \[\boxed{T’ \approx 993\,\text{N}}\] | The tension increases to approximately \(993\,\text{N}\) when the plate accelerates upward at \(0.05\,\text{m/s}^2\). |
Just ask: "Help me solve this problem."
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 |
|---|---|---|---|
| \( \text{AB} \) | \( 300 \) \( \text{mm} \) | \(\textbf{?}\) | \(\textbf{?}\) |
| \( \text{BC} \) | \( 600 \) \( \text{mm} \) | \(\textbf{?}\) | \( 1.2 \) \( \text{m/s} \) |
| \( \text{CD} \) | \(\textbf{?}\) | \( Q_{CD} = 2Q_{CE} \) \( \text{m}^3/\text{s} \) | \( 1.4 \) \( \text{m/s} \) |
| \( \text{CE} \) | \( 150 \) \( \text{mm} \) | \( Q_{CE} = 0.5Q_{CD} \) \( \text{m}^3/\text{s} \) | \(\textbf{?}\) |
A pump, submerged at the bottom of a well that is \( 35 \) \( \text{m} \) deep, is used to pump water uphill to a house that is \( 50 \) \( \text{m} \) above the top of the well, as shown to the right. The density of water is \( 1000 \) \( \text{kg/m}^3 \). All pressures are gauge pressures. Neglect the effects of friction, turbulence, and viscosity.
A Venturi meter is a device used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed \( v_2 \) through a horizontal section of pipe with a cross-sectional area \( A_2 = 542 \) \( \text{cm}^2 \). The gas has a density of \( 1.35 \) \( \text{kg/m}^3 \). The Venturi meter has a cross-sectional area of \( A_1 = 215 \) \( \text{cm}^2 \) and has been substituted for a section of the larger pipe. The pressure difference between the two sections \( P_2 – P_1 = 145 \) \( \text{Pa} \).
A spherical balloon of mass \( 226 \) \( \text{kg} \) is filled with helium gas until its volume is \( 325 \) \( \text{m}^3 \). Assume the density of air is \( 1.29 \) \( \text{kg/m}^3 \) and the density of helium is \( 0.179 \) \( \text{kg/m}^3 \).
A spherical balloon has a radius of \(7.15\) \(\text{m}\) and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of \(930\) \(\text{kg}\)?
Take the density of helium and air to be \(0.18\) \(\text{kg/m}^3\) and \(1.24\) \(\text{kg/m}^3\), respectively.
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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] |
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] |
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