0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (a): Determining the Spring Constant \(k\) | ||
| 1 | \[\Delta x_{\text{air}} = L_{\text{stretched}} – L_{\text{unstretched}}\] | Measure the length of the spring without the object and then with the object at rest in air. Their difference is the displacement \(\Delta x_{\text{air}}\). |
| 2 | \[mg = k\,\Delta x_{\text{air}}\] | At equilibrium in air, the downward gravitational force \(mg\) is balanced by the spring force \(k\,\Delta x_{\text{air}}\). |
| 3 | \[k = \frac{mg}{\Delta x_{\text{air}}}\] | Solve for \(k\) by algebraically rearranging the force equilibrium equation. |
| Alternative Method: One can also determine \(k\) by setting the mass into oscillation and using \(T = 2\pi\sqrt{\frac{m}{k}}\) to solve for \(k = \frac{4\pi^2m}{T^2}\), but the displacement method is straightforward using a metric ruler. | ||
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (b): Changes When the Object is Immersed | ||
| 1 | \[mg – F_{b} = k\,\Delta x_{\text{fluid}}\] | When the object is immersed in the fluid, it experiences an upward buoyant force \(F_{b}\). Therefore, the spring now only needs to balance the net force \(mg – F_{b}\), resulting in a smaller displacement \(\Delta x_{\text{fluid}}\) compared to \(\Delta x_{\text{air}}\). |
| 2 | \(\Delta x_{\text{fluid}} < \Delta x_{\text{air}}\) | The observed change is a decrease in the spring extension because the fluid’s buoyant force partially offsets the weight of the object. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (c): Experimental Determination of Fluid Density \(\rho\) | ||
| 1 | Measure \(\Delta x_{\text{air}}\) and \(\Delta x_{\text{fluid}}\). | Using the metric ruler, record the spring displacement when the object is in air and when it is immersed in the fluid. |
| 2 | Determine \(k = \frac{mg}{\Delta x_{\text{air}}}\) from Part (a). | This value of \(k\) is required for the next step of finding the buoyant force. |
| 3 | Measure the object’s mass \(m\) and use its known density \(D\) to find its volume \(V\) via \(V = \frac{m}{D}\). | By definition, density is mass per unit volume. Since \(D \gg \rho\), the object is practically incompressible and its volume can be calculated accurately. |
| 4 | Relate the buoyant force and the displaced fluid: \(F_{b} = \rho\,V\,g\). | According to Archimedes’ principle, the buoyant force equals the weight of the displaced fluid. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (d): Using Measurements to Calculate \(\rho\) | ||
| 1 | \(mg = k\,\Delta x_{\text{air}}\) | At equilibrium in air, the gravitational force is balanced by the spring force. |
| 2 | \(mg – k\,\Delta x_{\text{fluid}} = \rho\,V\,g\) | In the fluid, the buoyant force \(\rho\,V\,g\) reduces the effective force the spring must support. |
| 3 | Substitute \(k = \frac{mg}{\Delta x_{\text{air}}}\): \(mg – \frac{mg}{\Delta x_{\text{air}}}\,\Delta x_{\text{fluid}} = \rho\,V\,g\) | This substitution expresses the equation in terms of measurable quantities \(\Delta x_{\text{air}}\) and \(\Delta x_{\text{fluid}}\). |
| 4 | Divide by \(g\): \(m\Bigl(1- \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr) = \rho\,V\) | Simplify the equation by eliminating the gravitational acceleration \(g\), which appears on both sides. |
| 5 | Solve for \(\rho\): \(\displaystyle \rho = \frac{m \Bigl(1- \dfrac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr)}{V}\) | Isolate \(\rho\) to relate it directly to the measurements and the known mass and volume of the object. |
| 6 | Substitute \(V = \frac{m}{D}\): \(\displaystyle \rho = \frac{m \Bigl(1 – \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr)}{\frac{m}{D}} = D\Bigl(1 – \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr)\) | Since the object\’s density \(D\) and mass \(m\) give its volume, this substitution yields the final formula for the fluid density \(\rho\) in terms of \(D\) and the measured displacements. |
| 7 | \[\boxed{\rho = D \left(1 – \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\right)}\] | This is the explicit expression to calculate the fluid density based on the measured values. |
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?
An air mattress pump blows air above a beach ball at \( 8 \) \( \text{m/s} \). The air below the beach ball is moving at \( \approx 0 \) \( \text{m/s} \). Assuming the beach ball diameter is \( 0.1 \) \( \text{m} \), meaning the areas for the top \& bottom are each \( \approx 0.03 \) \( \text{m}^2 \), and the density of air is \( 1 \) \( \text{kg/m}^3 \), what is the lift force on the beach ball?
An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude \( A \) on a horizontal frictionless surface. Determine the following in terms of only the variable \( A \).
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?
Check the explanation for the complete solution. The following is a condensed version of the solution:
By continuing you (1) agree to our Terms of Sale and Terms of Use 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
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.
