| Derivation/Formula | Reasoning |
|---|---|
| \[0 = v_{iy}^2 – 2 g h\] | At maximum height the vertical speed is zero, so using \(v_y^2 = v_{iy}^2 – 2 g \Delta y\) with \(\Delta y = h\). |
| \[v_{iy} = \sqrt{2 g h}\] | Solve the previous equation for the initial vertical component \(v_{iy}\). |
| \[v_0 = \frac{v_{iy}}{\sin\theta}\] | Relate total launch speed to its vertical component using \(v_{iy} = v_0 \sin\theta\). |
| \[\boxed{v_0 \approx 2.24\,\text{m/s}}\] | Insert \(g = 9.80\,\text{m/s}^2\), \(h = 0.150\,\text{m}\), and \(\theta = 50^{\circ}\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[A_f = \pi r_f^2\] | Cross-sectional area of the fountain opening with radius \(r_f = 4.00\times10^{-3}\,\text{m}\). |
| \[Q = A_f v_0\] | Volume flow rate equals area times exit speed \(v_0\). |
| \[\boxed{Q \approx 1.13\times10^{-4}\,\text{m}^3/\text{s}}\] | Insert \(A_f = 5.03\times10^{-5}\,\text{m}^2\) and \(v_0 = 2.24\,\text{m/s}\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[A_p = \pi r_p^2\] | Area of feeder pipe with \(r_p = 7.00\times10^{-3}\,\text{m}\). |
| \[v_p = \frac{A_f}{A_p} v_0\] | Continuity equation \(A_f v_0 = A_p v_p\) gives pipe speed \(v_p\). |
| \[P_p + \tfrac{1}{2}\rho v_p^2 + \rho g y_p = P_0 + \tfrac{1}{2}\rho v_0^2 + \rho g y_0\] | Bernoulli between pipe point (subscript \(p\)) and nozzle (subscript \(0\)); nozzle pressure is atmospheric, taken as zero gauge. |
| \[P_p = \tfrac{1}{2}\rho (v_0^2 – v_p^2) + \rho g (y_0 – y_p)\] | Rearrange for gauge pressure; \(y_0 = 0\) at nozzle, \(y_p = -3.00\,\text{m}\). |
| \[\boxed{P_p \approx 3.16\times10^{4}\,\text{Pa}}\] | Insert \(\rho = 1.0\times10^{3}\,\text{kg/m}^3\), \(v_0 = 2.24\,\text{m/s}\), \(v_p = 0.731\,\text{m/s}\), and \(g = 9.80\,\text{m/s}^2\). |
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A student designs an experiment to determine the density of an unknown fluid. The student pours the fluid into a graduated cylinder and attaches an object to a force probe. The object has a density greater than the density of the fluid. The student partially submerges the object into the fluid and records both the volume of fluid displaced in the graduated cylinder and the reading on the force probe. The student then submerges the object further and, at each trial, records the new values of displaced volume and force probe reading until the object is fully submerged. The student constructs a graph of force probe reading (vertical axis) as a function of volume of fluid displaced (horizontal axis). Which of the following statements correctly describes how a feature of this graph is related to the density of the fluid?
A cannon fires projectiles on a flat range at a fixed speed but with variable angle. The maximum range of the cannon is \(L\). What is the range of the cannon when it fires at an angle of \(30^\circ\) above the horizontal? Ignore air resistance.
In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
A sample of an unknown material appears to weigh \( 285 \) \( \text{N} \) in air and \( 195 \) \( \text{N} \) when immersed in alcohol of specific gravity \( 0.700 \).
Two balls are launched at the same time from opposite sides of a \( 100 \) \( \text{m} \) wide and \(1000 ~\text{m}\) canyon. Ball A is launched at \( 20 \) \( \text{m/s} \) at \( 45^{\circ} \) from the left side. Ball B is launched at \( 20 \) \( \text{m/s} \) at \( 45^{\circ} \) from the right side.
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 37^{\circ} \). It lands on a platform that is \( 5.0 \) \( \text{m} \) above the launch height.
A projectile is launched at an angle of \( 30^{\circ} \) and hits a vertical wall \( 40 \) \( \text{m} \) away. After bouncing back horizontally, it lands \( 15 \) \( \text{m} \) behind the launch point. How high up on the wall did the projectile strike?
A ball is tossed straight up while the thrower is standing in a moving train car that is moving at a constant velocity. Neglecting air resistance, what is the path of the ball relative to the ground outside the train?
In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target? Explain without using equations.
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
\(2.24\,\text{m/s}\)
\(1.13\times10^{-4}\,\text{m}^3/\text{s}\)
\(3.16\times10^{4}\,\text{Pa}\)
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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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