0 attempts
0% avg
UBQ Credits
| 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. |
Just ask: "Help me solve this problem."
A cube of unknown material and uniform density floats in a container of water with \(60\%\) of its volume submerged. If this same cube were placed in a container of oil with density \(800\) \(\text{kg/m}^3\), what portion of the cube’s volume would be submerged while floating?
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?
When the button of a trash compactor is pushed, a force of \( 350 \) \( \text{N} \) pushes down on a \( 1.3 \) \( \text{cm}^2 \) input piston, creating a force of \( 22,076 \) \( \text{N} \) to crush the trash. What is the area of the piston that crushes the trash?
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.
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?
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
