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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."
An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
Two balls are launched at the same speed. Ball A is launched at an angle of \( 45^{\circ} \) and Ball B is launched at an angle of \( 60^{\circ} \). Which one reaches a higher point?
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?
Water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has the highest speed when it hits the ground? Ignore air resistance. Without using equations, explain your answer.
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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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