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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (a): Total Time to Reach 80 m | ||
| 1 | \[ t_{a} = \frac{v_x-0}{0.10} = \frac{2.0}{0.10} = 20\;\text{s} \] | Calculate the time required to reach a speed of \(2.0\;\text{m/s}\) from rest with a constant acceleration \(0.10\;\text{m/s}^2\). |
| 2 | \[ \Delta x_{a} = \frac{1}{2}(0.10)(20)^2 = 20\;\text{m} \] | Determine the displacement during the acceleration phase using the uniform acceleration equation. |
| 3 | \[ \Delta x_{c} = 80\;\text{m} – 20\;\text{m} = 60\;\text{m} \] | Find the remaining distance after the acceleration phase for which the rig travels at constant speed. |
| 4 | \[ t_{c} = \frac{60\;\text{m}}{2.0\;\text{m/s}} = 30\;\text{s} \] | Compute the time taken during the constant speed phase using \(\Delta x=t\,v_x\). |
| 5 | \[ T = t_{a} + t_{c} = 20\;\text{s} + 30\;\text{s} = 50\;\text{s} \] | Sum the two time intervals to get the total time to 80 m depth. |
| 6 | \[ \boxed{50\;\text{s}} \] | This is the total descent time to reach the maximum depth. |
| Part (b): Weight of the Water on the Top of the Bell | ||
| 1 | \[ \Delta P = \rho g h = 1025\;\text{kg/m}^3 \times 9.8\;\text{m/s}^2 \times 80\;\text{m} \] | Compute the hydrostatic pressure due to an 80 m water column (excluding the 1 atm inside the bell). |
| 2 | \[ \Delta P \approx 1025 \times 9.8 \times 80 \approx 803600\;\text{Pa} \] | Evaluate the product to obtain the pressure increase from the water column. |
| 3 | \[ F = \Delta P \times A = 803600\;\text{Pa} \times 9.0\;\text{m}^2 \] | Calculate the force (i.e. the weight of the water) on the bell’s top using its cross-sectional area. |
| 4 | \[ F \approx 7.2324 \times 10^6\;\text{N} \] | Multiply to get the force in newtons. |
| 5 | \[ \boxed{7.23 \times 10^6\;\text{N}} \] | This is the weight of the water exerted on the top of the bell at 80 m depth. |
| Part (c): Absolute Pressure on the Top of the Bell | ||
| 1 | \[ P_{\text{abs}} = P_{\text{atm}} + \rho g h \] | Add the atmospheric pressure at the surface to the hydrostatic pressure due to the 80 m water column. |
| 2 | \[ P_{\text{abs}} = 101325\;\text{Pa} + 803600\;\text{Pa} \] | Use \(101325\;\text{Pa}\) for 1 atm and the hydrostatic pressure found earlier. |
| 3 | \[ P_{\text{abs}} \approx 904925\;\text{Pa} \] | Sum the pressures to get the absolute pressure at depth. |
| 4 | \[ \boxed{9.05 \times 10^5\;\text{Pa}} \] | This is the absolute pressure on the top of the bell at 80 m depth. |
| Part (d): Minimum Force to Lift the Hatch | ||
| 1 | \[ A_{h} = \pi r^2 = \pi (0.25)^2 = \pi (0.0625) \approx 0.19635\;\text{m}^2 \] | Determine the area of the circular hatch with radius \(r = 0.25\;\text{m}\). |
| 2 | \[ F_{h} = \Delta P \times A_{h} = 803600\;\text{Pa} \times 0.19635\;\text{m}^2 \] | Calculate the net force acting on the hatch due to the pressure difference (\(\Delta P\)) computed earlier. |
| 3 | \[ F_{h} \approx 157800\;\text{N} \] | Multiply to find the minimum force necessary to overcome the water pressure on the hatch. |
| 4 | \[ \boxed{1.58 \times 10^5\;\text{N}} \] | This is the minimum force required to start lifting the hatch at the maximum depth. |
| Part (e): Reducing the Force to Open the Hatch | ||
| 1 | N/A | The force to open the hatch is given by the pressure difference multiplied by its area. To reduce this force, the net pressure difference must be decreased. |
| 2 | N/A | One effective method is to equalize the pressure on both sides of the hatch by increasing the internal pressure of the bell (for example, via a pressure equalization valve or controlled venting) so that it approaches the external hydrostatic pressure. |
| 3 | N/A | This reduces the differential pressure \(\Delta P\) acting on the hatch, thereby lowering the force required to open it. |
| 4 | N/A | Alternatively, decreasing the hatch area would also reduce the force, but modifying the pressure is generally more practical. |
| 5 | Answer: | To reduce the force, increase the bell’s internal pressure to nearly match the external pressure (or use a pressure equalization system), which minimizes the net pressure difference on the hatch. |
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In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
A beaker weighing \( 2.0 \) \( \text{N} \) is filled with \( 5.0 \times 10^{-3} \) \( \text{m}^3 \) of water. A rubber ball weighing \( 3.0 \) \( \text{N} \) is held entirely underwater by a massless string attached to the bottom of the beaker, as represented in the figure above. The tension in the string is \( 4.0 \) \( \text{N} \). The water fills the beaker to a depth of \( 0.20 \) \( \text{m} \). Water has a density of \( 1000 \) \( \text{kg/m}^3 \). The effects of atmospheric pressure may be neglected.
A trash compactor pushes down with a force of \( 500 \) \( \text{N} \) on a \( 3 \) \( \text{cm}^2 \) input piston, causing a force of \( 30,000 \) \( \text{N} \) to crush the trash. What is the area of the output piston that crushes the trash?
An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
A block of weight \( W \) is floating in water, and one-third of the block is above the surface of the water. Which of the following correctly describes the magnitude \( F \) of the force that the block exerts on the water and explains why \( F \) has that value?
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
The diagram above shows a hydraulic chamber with a spring \( (k_s = 1250 \, \text{N/m}) \) attached to the input piston and a rock of mass \( 55.2 \, \text{kg} \) resting on the output plunger. The input piston and output plunger are at about the same height, and each has negligible mass. The chamber is filled with water.
The figure shows a container filled with water to a depth \( d \). The container has a hole a distance \( y \) above its bottom, allowing water to exit with an initially horizontal velocity. Which of the following correctly predicts and explains how the speed of the water as it exits the hole would change if the distance \( y \) above the bottom of the container increased?
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?
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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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