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Part A – Minimum work required
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = 35\,\text{m} + 50\,\text{m} = 85\,\text{m}\] | The water is pumped from the bottom of the well (35 m deep) to a house 50 m above the well; hence the total vertical displacement is \(85\,\text{m}\). |
| 2 | \[W = \rho\,V\,g\,\Delta x\] | This formula gives the gravitational work (or potential energy gain) required to move a volume \(V\) of water with density \(\rho\) against gravity \(g\) through a vertical displacement \(\Delta x\). |
| 3 | \[W = (1000\,\text{kg/m}^3)(0.35\,\text{m}^3)(9.8\,\text{m/s}^2)(85\,\text{m})\] | Substitute the numerical values: water density \(\rho = 1000\,\text{kg/m}^3\), volume \(V = 0.35\,\text{m}^3\), gravitational acceleration \(g = 9.8\,\text{m/s}^2\), and displacement \(\Delta x = 85\,\text{m}\). |
| 4 | \[W \approx 2.92 \times 10^5\,\text{J}\] | Performing the multiplication gives \(W \approx 291550\,\text{J}\), which is rounded to \(2.92 \times 10^5\,\text{J}\). |
| 5 | \[\boxed{W \approx 2.92 \times 10^5\,\text{J}}\] | This is the minimum work required to pump the water used per day. |
Part B – Minimum power rating
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t = 2\,\text{hours} = 7200\,\text{s}\] | Convert the pumping time from hours to seconds for consistency in SI units. |
| 2 | \[P = \frac{W}{t}\] | Power is defined as the work done per unit time. |
| 3 | \[P = \frac{2.92 \times 10^5\,\text{J}}{7200\,\text{s}} \approx 40.5\,\text{W}\] | Substitute the work calculated in part (a) and the conversion for time to find the minimum power rating of the pump. |
| 4 | \[\boxed{P \approx 40.5\,\text{W}}\] | This represents the minimum power needed to pump the water within the specified time. |
Part C – Flow velocity
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[A_{\text{well}} = \frac{\pi (0.03)^2}{4}\] | Calculate the cross-sectional area of the pipe in the well (pipe diameter = 3.0 cm = 0.03 m). |
| 2 | \[Q = A_{\text{well}}\,v_i\] | The volumetric flow rate \(Q\) in the well is the product of the area and the well velocity \(v_i = 0.50\,\text{m/s}\). |
| 3 | \[A_{\text{house}} = \frac{\pi (0.0125)^2}{4}\] | Calculate the cross-sectional area of the pipe at the house (pipe diameter = 1.25 cm = 0.0125 m). |
| 4 | \[A_{\text{well}}\,v_i = A_{\text{house}}\,v_x \quad \Rightarrow \quad v_x = \frac{A_{\text{well}}}{A_{\text{house}}}\,v_i\] | Using mass continuity (volume flow rate is constant), relate the velocity in the well \(v_i\) to the velocity in the house \(v_x\) via their cross-sectional areas. |
| 5 | \[\frac{A_{\text{well}}}{A_{\text{house}}} = \left(\frac{0.03}{0.0125}\right)^2 = (2.4)^2 = 5.76\] | Since both areas involve the factor \(\pi/4\), the ratio simplifies to the square of the ratio of the diameters. |
| 6 | \[v_x = 5.76 \times 0.50 = 2.88\,\text{m/s}\] | Multiply the well velocity by the area ratio to obtain the flow velocity at the house’s faucet. |
| 7 | \[\boxed{v_x \approx 2.88\,\text{m/s}}\] | This is the calculated flow velocity when the faucet in the house is open. |
Part D – Calculating minimum pressure
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Energy per unit volume: } \Delta P = \rho g\,\Delta x + \frac{1}{2}\rho\,(v_x^2 – v_i^2)\] | This expression (derived from Bernoulli’s principle) represents the pressure energy per unit volume needed to overcome both the gravitational potential \(\rho g\,\Delta x\) and to provide the increase in kinetic energy from \(v_i\) (in the well) to \(v_x\) (at the faucet). |
| 2 | \[P_{\text{faucet, min}} = \rho g\,\Delta x + \frac{1}{2}\rho\,v_x^2\] | For minimum pressure at the faucet (assuming negligible initial kinetic energy \(v_i \approx 0\) or that its contribution is already included in the pump pressure), the faucet must supply at least the hydrostatic pressure plus the dynamic pressure necessary for the flow velocity \(v_x\). In practice, if the faucet discharges to the atmosphere, its gauge pressure is 0, so the pump must overcome this total pressure drop. |
| 3 | \[\boxed{P_{\text{min at faucet}} = \rho g\,\Delta x + \frac{1}{2}\rho\,v_x^2}\] | This is how one would calculate the minimum pressure required at the faucet: by summing the pressure needed to lift the water \(\rho g\,\Delta x\) and the pressure needed to accelerate it to \(v_x\) (i.e., \(\frac{1}{2}\rho\,v_x^2\)). |
| 4 | N/A | In summary, to find the minimum pressure at the faucet, you equate the pressure energy per unit volume provided by the pump with the sum of the gravitational and dynamic energy per unit volume required for the water to reach the faucet at velocity \(v_x\). This is essentially an algebraic application of Bernoulli’s principle under ideal (lossless) conditions. |
Just ask: "Help me solve this problem."
The box in the diagram is sliding to the right across a horizontal table, under the influence of the forces shown. Which force(s) is doing negative work on the box?
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
Diamond has a density of \( 3500 \) \( \text{kg/m}^3 \). During a physics lab, a diamond drops out of Virginia’s necklace and falls into her graduated cylinder filled with \( 5.00 \times 10^{-5} \) \( \text{m}^3 \) of water. This causes the water level to rise to the \( 5.05 \times 10^{-5} \) \( \text{m}^3 \) mark. What is the mass of Virginia’s diamond?
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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