0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ A_1 = \frac{\pi d_1^2}{4} \quad , \quad A_2 = \frac{\pi d_2^2}{4} \] | Calculate the cross-sectional areas of the pipe at street level and top floor using the area formula for a circle. Here, \(d_1 = 0.05\,m\) and \(d_2 = 0.028\,m\). |
| 2 | \[ A_1 v_i = A_2 v_x \] | Apply the continuity equation for incompressible flow which states that the volumetric flow rate must remain constant. |
| 3 | \[ v_x = \frac{A_1}{A_2} v_i = \left(\frac{d_1}{d_2}\right)^2 v_i \] | Simplify the expression by canceling the common factor \(\frac{\pi}{4}\) in the area formulas, showing that the velocity scales as the square of the diameter ratio. |
| 4 | \[ v_x = \left(\frac{0.05}{0.028}\right)^2 (0.78) \approx 2.48\,m/s \] | Substitute the given values: \(d_1 = 0.05\,m\), \(d_2 = 0.028\,m\), and \(v_i = 0.78\,m/s\). The ratio \(\left(\frac{0.05}{0.028}\right)^2 \) is approximately 3.188; multiplying by 0.78 yields \(v_x \approx 2.48\,m/s\). |
| 5 | \[ \boxed{v_x \approx 2.48\, m/s} \] | This is the final computed flow velocity at the top floor of the building. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \frac{P_1}{\rho} + \frac{v_i^2}{2} + g y_1 = \frac{P_2}{\rho} + \frac{v_x^2}{2} + g y_2 \] | Write Bernoulli’s equation between the street level (point 1) and the top floor (point 2). Here, \(y_1 = 0\) and \(y_2 = 16\,m\). |
| 2 | \[ \frac{P_2}{\rho} = \frac{P_1}{\rho} + \frac{v_i^2 – v_x^2}{2} – g \Delta y \] | Rearrange the equation to solve for \(P_2\), where \(\Delta y = y_2 – y_1 = 16\,m\). |
| 3 | \[ P_2 = P_1 + \rho \left(\frac{v_i^2 – v_x^2}{2} – g \Delta y \right) \] | Multiply both sides by the density \(\rho\) to isolate \(P_2\). |
| 4 | \[ P_1 = 3.8\,atm = 3.8 \times 101325 \approx 385035\,Pa \] | Convert the gauge pressure at street level from atmospheres to Pascals using \(1\,atm \approx 101325\,Pa\). |
| 5 | \[ \frac{v_i^2 – v_x^2}{2} = \frac{0.78^2 – 2.48^2}{2} \approx \frac{0.6084 – 6.1504}{2} \approx -2.77\, m^2/s^2 \] | Calculate the difference in kinetic energy per unit mass between the two points. |
| 6 | \[ g \Delta y = 9.8 \times 16 \approx 156.8\, m^2/s^2 \] | Compute the gravitational potential energy change per unit mass over a height of \(16\,m\). |
| 7 | \[ \rho \left( \frac{v_i^2 – v_x^2}{2} – g \Delta y \right) \approx 1000 \left(-2.77 – 156.8\right) \approx -159570\,Pa \] | Combine the kinetic and potential terms multiplied by the density \(\rho = 1000\,kg/m^3\) to find the pressure change. |
| 8 | \[ P_2 \approx 385035 – 159570 \approx 225465\,Pa \] | Subtract the pressure drop from the initial gauge pressure to get the gauge pressure at the top floor. |
| 9 | \[ \boxed{P_2 \approx 2.25 \times 10^5\,Pa} \] | This is the final gauge pressure of the water in the pipe on the top floor. |
Just ask: "Help me solve this problem."
A car is going through a dip in the road whose curvature approximates a circle of radius \( 200 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 20\% \) more than their normal weight \( (1.2\,W) \)?
The occupants of a car traveling at a speed of \( 30 \) \( \text{m/s} \) note that on a particular part of a road their apparent weight is \( 15\% \) higher than their weight when driving on a flat road.
A ball is attached to the end of a string. It is swung in a vertical circle of radius \( 0.33 \) \( \text{m} \). What is the minimum velocity that the ball must have to make it around the circle?
A ball is attached to the end of a string. It is swung in a vertical circle of radius \( 2.5 \) \( \text{m} \). What is the minimum velocity that the ball must have to make it around the circle?
A car is going over the top of a hill whose curvature approximates a circle of radius \( 350 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 10\% \) less than their normal weight?
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted the ultimate A.P Physics 1 course that simplifies everything so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
