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| 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. |
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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?
A cube of side length \( s \) rests on the bottom surface of a container of fluid. The fluid is at a height \( y \) above the bottom of the tank. The fluid has density \( \rho \) and the atmospheric pressure is \( P_{\text{atm}} \).
Which of the following expressions is equal to the absolute pressure exerted by the fluid on the top surface of the cube?
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.
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 fountain with an opening of radius \( 0.015 \) \( \text{m} \) shoots a stream of water vertically from ground level at \( 6.0 \) \( \text{m/s} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
Two blocks of the same size are floating in a container of water. The first block is submerged \( 80\% \) while the second block is submerged by \( 20\% \) beneath the water. Which of the following is a correct statement about the two blocks?
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 \( 0.179 \) \( \text{kg/m}^3 \).
A fluid flows through the two sections of cylindrical pipe shown in the figure. The narrow section of the pipe has radius \( R \) and the wide section has radius \( 2R \). What is the ratio of the fluid’s speed in the wide section of pipe to its speed in the narrow section of pipe, \( \frac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
A person is standing on a railroad station platform when a high-speed train passes by. The person will tend to be
The radius of the left piston is \( 0.12 \) \( \text{m} \) and the radius of the right piston is \( 0.65 \) \( \text{m} \). If \( f \) were raised by \( 14 \) \( \text{N} \), how much would \( F \) need to be increased to maintain equilibrium?
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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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