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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[W_{\text{water}} = \rho g V = (1000\, \text{kg/m}^3)(9.8\, \text{m/s}^2)(5.0\times10^{-3}\, \text{m}^3)=4.9\times10^{1}\, \text{N}\] | Use \( W = \rho g V \) to convert the given water volume into its weight. |
| 2 | \[W_{\text{total}} = 2.0\, \text{N} + 4.9\times10^{1}\, \text{N} + 3.0\, \text{N}=\boxed{5.4\times10^{1}\, \text{N}}\] | Add weights of the beaker, water, and ball. Tension is internal, so it does not affect total weight. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Forces:}\; W=3.0\,\text{N (down)},\; T=4.0\,\text{N (down)},\; B\,\text{(up)}\] | Identify all forces acting on the ball: its weight \(W\), the downward tension \(T\), and the upward buoyant force \(B\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[B = W + T = 3.0\, \text{N} + 4.0\, \text{N} = \boxed{7.0\, \text{N}}\] | The ball is in static equilibrium, so upward buoyant force equals the sum of downward forces. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[P = \rho g h = (1000\, \text{kg/m}^3)(9.8\, \text{m/s}^2)(0.20\, \text{m}) = \boxed{2.0\times10^{3}\, \text{Pa}}\] | Use hydrostatic relation \(P = \rho g h\) at the given depth \(h=0.20\,\text{m}\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Initial displaced weight} = B = 7.0\,\text{N} > W_{\text{ball}} = 3.0\,\text{N}\] | While submerged, the ball displaces water whose weight equals the buoyant force (7 N). |
| 2 | \[\text{Floating displacement} = W_{\text{ball}} = 3.0\,\text{N}\] | Once free, a floating object displaces only its own weight in fluid. |
| 3 | \[\boxed{\text{Water level lower}}\] | The displaced volume (and thus water level) decreases from that producing 7 N to that producing 3 N, so the level drops. |
Just ask: "Help me solve this problem."
Water flowing in a horizontal pipe speeds up as it goes from a section with a large diameter to a section with a small diameter. Which of the following can explain why the speed of the water increases?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
An ideal fluid flows through a pipe with radius \( Q \) and flow speed \( V \). If the pipe splits up into three separate paths, each with radius \( \frac{Q}{2} \), what is the flow speed through each of the paths?
Three identical reservoirs, \(A\), \(B\), and \(C\), are represented above, each with a small pipe where water exits horizontally. The pipes are set at the same height above a pool of water. The water in the reservoirs is kept at the levels shown. Which of the following correctly ranks the horizontal distances \( d \) that the streams of water travel before hitting the surface of the pool?
Johnny the auto mechanic is raising a \( 1200 \) \( \text{kg} \) car on her hydraulic lift so that she can work underneath. If the area of the input piston is \( 12 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
\(5.4\times10^{1}\,\text{N}\)
\(W\,\text{down},\;T\,\text{down},\;B\,\text{up}\)
\(7.0\,\text{N}\)
\(2.0\times10^{3}\,\text{Pa}\)
\(\text{lower}\)
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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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