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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ h = \frac{(v_i \sin 50^\circ)^2}{2g} \] | This is the formula for the maximum height reached by a projectile in terms of the initial speed \(v_i\) and launch angle \(50^\circ\). |
| 2 | \[ 0.150 = \frac{(v_i \sin 50^\circ)^2}{2 \times 9.8} \] | Substitute the given maximum height \(h = 0.150\,m\) and \(g = 9.8\,m/s^2\) into the formula. |
| 3 | \[ (v_i \sin 50^\circ)^2 = 0.150 \times 19.6 \] | Multiply both sides by \(2 \times 9.8 = 19.6\) to isolate the squared term. |
| 4 | \[ (v_i \sin 50^\circ)^2 = 2.94 \] | Calculating \(0.150 \times 19.6\) gives \(2.94\). |
| 5 | \[ v_i \sin 50^\circ = \sqrt{2.94} \] | Apply the square root to both sides to solve for \(v_i \sin 50^\circ\). |
| 6 | \[ v_i = \frac{\sqrt{2.94}}{\sin 50^\circ} \] | Solve for the initial speed \(v_i\) by dividing by \(\sin 50^\circ\). |
| 7 | \[ v_i \approx \frac{1.714}{0.766} \approx 2.24\,m/s \] | Using \(\sqrt{2.94} \approx 1.714\) and \(\sin 50^\circ \approx 0.766\), we calculate \(v_i \approx 2.24\,m/s\). |
| 8 | \[ \boxed{v_i \approx 2.24\,m/s} \] | This is the final speed at which the water leaves the fountain. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A = \pi r^2 \] | Calculate the cross-sectional area of the fountain’s exit hole, where \(r = 4.00 \times 10^{-3}\,m\). |
| 2 | \[ A = \pi (4.00 \times 10^{-3})^2 = \pi (16.00 \times 10^{-6}) \approx 5.03 \times 10^{-5}\,m^2 \] | Squaring the radius and multiplying by \(\pi\) gives the area. |
| 3 | \[ Q = A \times v_i \] | The volume rate of flow \(Q\) is found by multiplying the exit hole area by the speed \(v_i\) from part (a). |
| 4 | \[ Q \approx 5.03 \times 10^{-5}\,m^2 \times 2.24\,m/s \approx 1.13 \times 10^{-4}\,m^3/s \] | Multiply the area by the speed to obtain the volumetric flow rate. |
| 5 | \[ \boxed{Q \approx 1.13 \times 10^{-4}\,m^3/s} \] | This is the final volume rate of flow of the water. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ P = \rho g h \] | The gauge pressure due to a fluid column is given by \(P = \rho g h\), where \(h\) is the vertical distance below the fountain’s opening. |
| 2 | \[ P = (1.0 \times 10^3\,kg/m^3)(9.8\,m/s^2)(3.00\,m) \] | Substitute the density \(\rho = 1.0 \times 10^3\,kg/m^3\), gravitational acceleration \(g = 9.8\,m/s^2\), and height \(h = 3.00\,m\) into the formula. |
| 3 | \[ P = 29,400\,Pa \] | Multiplying gives the gauge pressure at the point in the feeder pipe. |
| 4 | \[ \boxed{P \approx 2.94 \times 10^4\,Pa} \] | This is the final gauge pressure in the feeder pipe at 3.00 m below the fountain’s opening. |
Just ask: "Help me solve this problem."
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
A ball of mass \( M \) is attached to a string of length \( L \). It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is \( 3 \) times the weight of the ball. Give all answers in terms of \( M \), \( L \), and \( g \).
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?
Which of the following statements is an expression of the equation of continuity?
Suppose the water at the top of Niagara Falls has a horizontal speed of \( 2.7 \, \text{m/s} \) just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a \( 75^\circ \) angle below the horizontal?
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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] |
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] |
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