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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."
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 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.
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?
Two objects labeled K and L have equal mass but densities \( 0.95D_o \) and \( D_o \), respectively. Each of these objects floats after being thrown into a deep swimming pool. Which is true about the buoyant forces acting on these objects?
In the lab, a student is given a glass beaker filled with water with an ice cube of mass \( m \) and volume \( V_c \) floating in it.
The downward force of gravity on the ice cube has magnitude \( F_g \). The student pushes down on the ice cube with a force of magnitude \( F_P \) so that the cube is totally submerged. The water then exerts an upward buoyant force on the ice cube of magnitude \( F_B \). Which of the following is an expression for the magnitude of the acceleration of the ice cube when it is released?
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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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