| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ F_{\text{out}} = m \cdot g \] | Calculate the weight of the car where \( m = 1200 \) kg and \( g \approx 9.8 \) m/s\(^{2}\). |
| 2 | \[ F_{\text{out}} = 1200 \times 9.8 = 11760 \, \text{N} \] | Substitute the values into the weight formula to find the force on the output piston. |
| 3 | \[ \frac{F_{\text{in}}}{A_{\text{in}}} = \frac{F_{\text{out}}}{A_{\text{out}}} \] | Apply Pascal’s Principle which states that the pressure is constant throughout the hydraulic system. |
| 4 | \[ F_{\text{in}} = \frac{A_{\text{in}}}{A_{\text{out}}} \cdot F_{\text{out}} \] | Rearrange the pressure equality to solve for the input force \( F_{\text{in}} \). |
| 5 | \[ F_{\text{in}} = \frac{12.0}{700} \times 11760 \] | Substitute the areas \( A_{\text{in}} = 12.0 \) cm\(^{2}\) and \( A_{\text{out}} = 700 \) cm\(^{2}\) along with \( F_{\text{out}} \) into the formula. Note: Since the areas are given in the same units, the ratio is unitless. |
| 6 | \[ F_{\text{in}} = \left( \frac{12.0}{700} \right) \times 11760 \approx 201.6 \, \text{N} \] | Perform the multiplication to calculate the force required on the input piston. |
| 7 | \[ \boxed{F_{\text{in}} \approx 202 \, \text{N}} \] | Round the result to three significant figures to get the final answer. |
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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 \( 1000 \) \( \text{kg/m}^3 \).
Water of density \( \rho \) flows through the section of circular pipe shown in the figure. At Point A, where the diameter of the pipe is \( D \), the water has a pressure \( P_0 \) and velocity \( v_0 \). Point B is located a vertical distance \( H \) above Point A in a section of the pipe that has diameter \( 2D \).
Which of the following expressions is equal to the pressure of the water at Point B?
A student places a wooden block of mass \( m \) in a container of water. The block floats with half of its volume above the surface of the water. The student then begins to stack small objects on top of the block until the wooden block is completely submerged but none of the objects stacked on top are submerged. What is the buoyant force acting on the block now?
Find the approximate minimum mass needed for a spherical ball with a \(40\) \(\text{cm}\) radius to sink in a liquid of density \(1.4 \times 10^3\) \(\text{kg/m}^3\). Use \(9.8 \text{m/s}^2\) for \(g\).
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?
In the lab, a student is given a glass beaker filled with water with an ice cube of mass \( m \) and volume \( V_c \) floating in it.
The downward force of gravity on the ice cube has magnitude \( F_g \). The student pushes down on the ice cube with a force of magnitude \( F_P \) so that the cube is totally submerged. The water then exerts an upward buoyant force on the ice cube of magnitude \( F_B \). Which of the following is an expression for the magnitude of the acceleration of the ice cube when it is released?
Diamond has a density of \( 3500 \) \( \text{kg/m}^3 \). During a physics lab, a diamond drops out of Virginia’s necklace and falls into her graduated cylinder filled with \( 5.00 \times 10^{-5} \) \( \text{m}^3 \) of water. This causes the water level to rise to the \( 5.05 \times 10^{-5} \) \( \text{m}^3 \) mark. What is the mass of Virginia’s diamond?
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 pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
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?
\(202 \text{ N}\)
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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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