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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."
Find the approximate minimum mass needed for a spherical ball with a \(40\) \(\text{cm}\) radius to sink in a liquid of density \(1.4 \times 10^3\) \(\text{kg/m}^3\). Use \(9.8 \text{m/s}^2\) for \(g\).
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 experimental diving rig is lowered from rest at the ocean’s surface and reaches a maximum depth of \(80\) \(\text{m}\). Initially it accelerates downward at a rate of \(0.10\) \(\text{m/s}^2\) until it reaches a speed of \(2.0\) \(\text{m/s}\), which then remains constant. During the descent, the pressure inside the bell remains constant at \(1\) atmosphere. The top of the bell has a cross-sectional area \(A = 9.0\) \(\text{m}^2\). The density of seawater is \(1025\) \(\text{kg/m}^3\).
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?
Rex, an auto mechanic, is raising a \( 1200 \) \( \text{kg} \) car on his hydraulic lift so that he can work underneath. If the area of the input piston is \( 12.0 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
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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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