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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[A_1v_1 = A_2v_2\] | Apply the principle of continuity, which states that for an incompressible fluid, the mass flow rate must be constant. This implies that the product of the cross-sectional area and the velocity is constant along the flow path. |
| 2 | \[\frac{v_2}{v_1} = \frac{A_1}{A_2} = \left(\frac{d_1}{d_2}\right)^2 = 100\] | Given that the diameter of the hose is \(10\) times that of the nozzle, the area ratio \(\left(\frac{A_1}{A_2}\right)\) is \(10^2 = 100\). Therefore, \(v_2 = 100 \times v_1\). |
| 3 | \[v_2 = 100 \times 0.4 \, \text{m/s} = 40 \, \text{m/s}\] | Substitute \(v_1 = 0.4 \, \text{m/s}\) into the equation to find \(v_2\). This is the velocity of water at the nozzle. |
| 4 | \[P_1 + \frac{1}{2}\rho v_1^2 + \rho gy_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gy_2\] | Apply Bernoulli’s Equation considering points at the pump (Point 1) and at the nozzle (Point 2). Pressure, kinetic energy per unit volume, and potential energy per unit volume are balanced between the two points. |
| 5 | \[P_1 + \frac{1}{2}\rho (0.4)^2 + 0 = P_{\text{atm}} + \frac{1}{2}\rho (40)^2 + \rho g(1)\] | Substitute known values: \(v_1 = 0.4 \, \text{m/s}\), \(y_1 = 0\), \(y_2 = 1\), and \(v_2 = 40 \, \text{m/s}\). At Point 2, pressure equals atmospheric pressure \(P_{\text{atm}}\). |
| 6 | \[P_1 = P_{\text{atm}} + \rho g + \frac{1}{2}\rho (40)^2 – \frac{1}{2}\rho (0.4)^2\] | Reorganize the equation to express the pressure at the pump, \(P_1\), in terms of atmospheric pressure and other known quantities. |
| 7 | \[P_1 – P_{\text{atm}} = \rho g + \frac{1}{2}\rho ((40)^2 – (0.4)^2)\] | Calculate the pressure difference between the pump and the atmosphere. |
| 8 | \[P_1 – P_{\text{atm}} = 1000 \times 9.8 + \frac{1}{2} \times 1000 ((40)^2 – (0.4)^2)\] | Use \(\rho = 1000 \, \text{kg/m}^3\) for the density of water and \(g = 9.8 \, \text{m/s}^2\) for gravitational acceleration. Calculate the individual energy terms in the equation. |
| 9 | \[P_1 – P_{\text{atm}} = 9800 + \left( \frac{1}{2} \right) 1000 \times (1600 – 0.16)\] | Substitute and simplify the calculation for kinetic and potential energies. |
| 10 | \[P_1 – P_{\text{atm}} = 9800 + 800000\] | Complete the calculations: \((1600 – 0.16) = 1599.84\). Therefore, \(\frac{1}{2} \times 1000 \times 1599.84 = 799920\) Pa. |
| 11 | \[P_1 – P_{\text{atm}} = 809800 \, \text{Pa}\] | Convert the final result to kilopascals \( \text{kPa} \) (1 \(\text{kPa} = 1000 \text{Pa} \)). Box the final answer. |
| 12 | \[ \boxed{810 \, \text{kPa}} \] | The result shows the pressure difference between the pump and the atmospheric pressure. The correct multiple-choice answer is \( (d) \, 810 \, \text{kPa} \). |
Just ask: "Help me solve this problem."
The figure shows a horizontal pipe with sections with different cross-sectional areas. Small tubes extend from the top of each section. The cross-sectional area of the pipe at location C is half that at A, and the areas at A and D are the same. Water flows in the pipe from left to right. Which of the following correctly ranks the height \( h \) of the water in the tubes above the labeled locations?
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?
The diagram above shows a hydraulic chamber with a spring \( (k_s = 1250 \, \text{N/m}) \) attached to the input piston and a rock of mass \( 55.2 \, \text{kg} \) resting on the output plunger. The input piston and output plunger are at about the same height, and each has negligible mass. The chamber is filled with water.
Find the approximate minimum mass needed for a spherical ball with a \(40\) \(\text{cm}\) radius to sink in a liquid of density \(1.4 \times 10^3\) \(\text{kg/m}^3\). Use \(9.8 \text{m/s}^2\) for \(g\).
Rex, an auto mechanic, is raising a \( 1200 \) \( \text{kg} \) car on his hydraulic lift so that he can work underneath. If the area of the input piston is \( 12.0 \) \( \text{cm}^2 \), 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?
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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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