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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(Q_{\text{initial}} = A_{\text{initial}} \cdot V_{\text{initial}}\) | The flow rate through the initial pipe is given by the product of the cross-sectional area and the flow speed. |
| 2 | \(A_{\text{initial}} = \pi Q^2\) | The cross-sectional area of the initial pipe is calculated using the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius of the initial pipe. |
| 3 | \(Q_{\text{initial}} = \pi Q^2 \cdot V\) | Substitute the expression for \(A_{\text{initial}}\) into the flow rate equation to obtain the initial flow rate. |
| 4 | \(Q_{\text{final\ total}} = 3 \cdot A_{\text{path}} \cdot V_{\text{path}}\) | Since the flow is split equally into three paths, the total flow rate through the paths is three times the flow rate through one path. |
| 5 | \(A_{\text{path}} = \pi \left(\frac{Q}{2}\right)^2 = \frac{\pi Q^2}{4}\) | Calculate the cross-sectional area of one path using its radius, \( \frac{Q}{2} \). |
| 6 | \(Q_{\text{final\ total}} = 3 \cdot \frac{\pi Q^2}{4} \cdot V_{\text{path}}\) | Substitute \(A_{\text{path}}\) into the flow rate equation for the paths. |
| 7 | \(\pi Q^2 \cdot V = 3 \cdot \frac{\pi Q^2}{4} \cdot V_{\text{path}}\) | Apply the conservation of mass, equating the initial flow rate to the total flow rate of the split paths. |
| 8 | \(V_{\text{path}} = \frac{4}{3} V\) | Solve for the flow speed through each path, showing that it is \( \frac{4}{3} \) times the initial flow speed, matching option (c). |
| 9 | Correct answer: (c) \( \boxed{\frac{4}{3} V} \) | The calculation confirms that the flow speed through each path is \( \frac{4}{3} \) times the original speed. |
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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 \( 1000 \) \( \text{kg/m}^3 \).
A pump, submerged at the bottom of a well that is \( 35 \) \( \text{m} \) deep, is used to pump water uphill to a house that is \( 50 \) \( \text{m} \) above the top of the well, as shown to the right. The density of water is \( 1000 \) \( \text{kg/m}^3 \). All pressures are gauge pressures. Neglect the effects of friction, turbulence, and viscosity.
A \(2\)-N force is used to push a small piston \(10\) \(\text{cm}\) downward in a simple hydraulic machine. If the opposite large piston rises by \(0.5\) \(\text{cm}\), what is the maximum weight the large piston can lift?
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 beaker weighing \( 2.0 \) \( \text{N} \) is filled with \( 5.0 \times 10^{-3} \) \( \text{m}^3 \) of water. A rubber ball weighing \( 3.0 \) \( \text{N} \) is held entirely underwater by a massless string attached to the bottom of the beaker, as represented in the figure above. The tension in the string is \( 4.0 \) \( \text{N} \). The water fills the beaker to a depth of \( 0.20 \) \( \text{m} \). Water has a density of \( 1000 \) \( \text{kg/m}^3 \). The effects of atmospheric pressure may be neglected.
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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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