| 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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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 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 \( 0.179 \) \( \text{kg/m}^3 \).
Suppose we wish to make a neutrally buoyant hollow sphere out of titanium (\(\rho = 4500 \text{kg/m}^3\)). If the sphere has an outer radius of \( 1.5 \) \( \text{m} \), what must be its inner radius?
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.
Caleb is filling up water balloons for the Physics Olympics balloon toss competition. Caleb sets a \( 0.50 \text{-kg} \) spherical water balloon on the kitchen table and notices that the bottom of the balloon flattens until the pressure on the bottom is reduced to \( 630 \frac{\text{N}}{\text{m}^2} \). What is the area of the flat spot on the bottom of the balloon?
You have a giant cask of water with a spigot some height below the water surface. The surface of the water, which is essentially at rest, is exposed to atmosphere (\( \approx 10^5 \text{Pa} \)). The water density is \( \approx 1000 \text{kg/m}^3 \). The water pours out of the spigot at \( 3 \text{m/s} \). How far below the water surface is the spigot positioned?
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?
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?
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 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.
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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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