| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ F_\text{output} = mg \] | Calculate the force due to the mass of the rock. Here, \( m = 55.2 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \). |
| 2 | \[ F_\text{output} = 55.2 \times 9.81 \] | Substitute values to find the force on the output plunger. |
| 3 | \[ F_\text{output} = 541.212 \, \text{N} \] | The force exerted by the rock on the output plunger. |
| 4 | \[ \frac{F_\text{input}}{A_\text{input}} = \frac{F_\text{output}}{A_\text{output}} \] | Use Pascal’s principle, which states that pressure is transmitted undiminished in an enclosed static fluid. |
| 5 | \[ F_\text{input} = \frac{F_\text{output} \times A_\text{input}}{A_\text{output}} \] | Rearrange to solve for the input force needed for equilibrium. |
| 6 | \[ F_\text{input} = \frac{541.212 \times 15}{65} \] | Substitute the area values: \( A_\text{input} = 15 \, \text{cm}^2 \) and \( A_\text{output} = 65 \, \text{cm}^2 \). |
| 7 | \[ F_\text{input} = 124.843 \, \text{N} \] | Calculate the force exerted on the input piston necessary for equilibrium. |
| 8 | \[ 124.843 = k_s \Delta x \] | Relate the input force to the spring constant \( k_s = 1250 \, \text{N/m} \) and the compression \( \Delta x \). |
| 9 | \[ \Delta x = \frac{124.843}{1250} \] | Solve for the compression of the spring. |
| 10 | \[ \Delta x = 0.0999 \, \text{m} \] | Convert the compression to meters. |
| 11 | \[ \boxed{9.99 \, \text{cm}} \] | Convert to centimeters and box the final answer. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_\text{input} \Delta y_\text{input} = A_\text{output} \Delta y_\text{output} \] | Use the principle of conservation of volume in the hydraulic system. |
| 2 | \[ 15 \times 22.0 = 65 \times \Delta y_\text{output} \] | Substitute \( \Delta y_\text{input} = 22.0 \, \text{cm} \) and the areas. |
| 3 | \[ 330 = 65 \times \Delta y_\text{output} \] | Calculate the product of the input area and the distance. |
| 4 | \[ \Delta y_\text{output} = \frac{330}{65} \] | Solve for the rise in the output plunger’s height. |
| 5 | \[ \Delta y_\text{output} = 5.077 \, \text{cm} \] | The final rise in the output plunger. |
| 6 | \[ \boxed{5.08 \, \text{cm}} \] | Box the final answer after rounding to two decimal places. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ P = P_0 + \rho g h \] | The absolute pressure at a depth \( h \) is given by this equation, where \( P_0 \) is atmospheric pressure. |
| 2 | \[ P = 101325 + 1000 \times 9.81 \times 0.85 \] | Substitute \( P_0 = 101325 \, \text{Pa} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.81 \, \text{m/s}^2 \), and the height \( h = 0.85 \, \text{m} \). |
| 3 | \[ P = 101325 + 8338.5 \] | Calculate the pressure contribution from the water column. |
| 4 | \[ P = 109663.5 \, \text{Pa} \] | Calculate the total absolute pressure at the bottom of the chamber. |
| 5 | \[ \boxed{109664 \, \text{Pa}} \] | Box the final answer after rounding to the nearest Pascal. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
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?
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?
Wanda watches the fish in her fish tank and notices that the angelfish like to feed at the water’s surface, while the catfish feed \( 0.300 \) \( \text{m} \) below at the bottom of the tank. If the average density of the water in the tank is \( 1000\) \( \text{kg/m}^3 \), what is the absolute pressure on the catfish?
A person is standing on a railroad station platform when a high-speed train passes by. The person will tend to be
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.
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?
Two paper cups are suspended by strings and hung near each other. They are separated by about \( 10 \) \( \text{cm} \). Explain what happens to the cups when you blow air between them. Hint: Do they remain still, moves away from each other or move towards each other?
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 \(1.025 \times 10^3 \text{ kg/m}^3\).
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.
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
By continuing you (1) agree to our Terms of Use and Terms of Sale 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] |
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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 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.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
