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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 rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?
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?
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.
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
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?
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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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