| 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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Three identical reservoirs, \(A\), \(B\), and \(C\), are represented above, each with a small pipe where water exits horizontally. The pipes are set at the same height above a pool of water. The water in the reservoirs is kept at the levels shown. Which of the following correctly ranks the horizontal distances \( d \) that the streams of water travel before hitting the surface of the pool?
The large piston in a hydraulic lift has a radius of \( 250 \) \( \text{cm}^2 \). What force must be applied to the small piston with a radius of \( 25 \) \( \text{cm}^2 \) in order to raise a car of mass \( 1500 \) \( \text{kg} \)?
An ideal fluid flows through a pipe with radius \( Q \) and flow speed \( V \). If the pipe splits up into three separate paths, each with radius \( \frac{Q}{2} \), what is the flow speed through each of the paths?
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?
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
A geologist suspects that her rock specimen is hollow, so she weighs the specimen in both air and water. When completely submerged, the rock weighs twice as much in air as it does in water.
The drawing above shows a spherical reservoir that contains \( 455,000 \) \( \text{kg} \) of water when full. The reservoir is vented to the atmosphere at the top. Assuming the reservoir is full and the diameter of the reservoir is much larger than any of the pipes on the ground.
Nancy is using a turkey baster (a kitchen tool with a rubber bulb on one end and a tube on the other) to collect juices from a roasting turkey. When she squeezes and then releases the rubber bulb, it creates suction with a pressure of \( 99{,}800 \) \( \text{Pa} \). This suction causes the turkey juice to rise \( 9 \) \( \text{cm} \) up the tube. Based on this information, what is the density of the turkey juice?
A diver descends from a salvage ship to the ocean floor at a depth of \(35 \text{ m}\) below the surface. The density of ocean water is \(1.025 \times 10^3 \text{ kg/m}^3\).
The figure above shows a portion of a conduit for water, one with rectangular cross sections. If the flow speed at the top is \( v \), what is the flow speed at the bottom?
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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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