| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( W_{\text{block}} = F_{\text{buoyant}} \) | Since the block is floating, it is in equilibrium, meaning the upward buoyant force equals the downward gravitational force (weight of the block). This is due to Archimedes’ principle which states that the buoyant force is equal to the weight of the fluid displaced by the submerged part of the block. |
| 2 | \( V_{\text{submerged}} = \frac{2}{3}V_{\text{total}} \) | Two-thirds of the block is submerged in the water, which is given in the problem statement. |
| 3 | \( F_{\text{buoyant}} = \rho_{\text{water}} \cdot g \cdot V_{\text{submerged}} \) | The buoyant force is equal to the product of the density of the water, gravitational acceleration, and the volume of the submerged part of the block. |
| 4 | \( F_{\text{buoyant}} = \rho_{\text{water}} \cdot g \cdot \left(\frac{2}{3}V_{\text{total}}\right) \) | Substituting for \( V_{\text{submerged}} \) from the reasoning in step 2. |
| 5 | \( F_{\text{block\ on\ water}} = W = \rho_{\text{water}} \cdot g \cdot \left(\frac{2}{3}V_{\text{total}}\right) \) | According to Newton’s third law, the force that the block exerts on the water is equal in magnitude and opposite in direction to the buoyant force. Therefore, \( F = W \). |
| 6 | (c) \( F = W \), as the block is at rest so the net force must be zero. The water must exert an upward force \( F = W \) to balance the block’s weight, and the block exerts an equal force on the water. | This option (d) is correct because it correctly applies the principle of buoyancy and equilibrium conditions without misinterpreting the force exerted based on the submerged volume alone. |
Explanation of Incorrect Answers:
– Option (a) incorrectly assumes the contact surface area determines the force rather than the weight balance.
– Option (b) misinterprets the source of the force by focusing solely on the submerged portion rather than considering equilibrium.
– Option (d) oversimplifies by suggesting all objects exert forces equal to their weight in any contact scenario, which overlooks fluid dynamics and equilibrium principles.
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A student designs an experiment to determine the density of an unknown fluid. The student pours the fluid into a graduated cylinder and attaches an object to a force probe. The object has a density greater than the density of the fluid. The student partially submerges the object into the fluid and records both the volume of fluid displaced in the graduated cylinder and the reading on the force probe. The student then submerges the object further and, at each trial, records the new values of displaced volume and force probe reading until the object is fully submerged. The student constructs a graph of force probe reading (vertical axis) as a function of volume of fluid displaced (horizontal axis). Which of the following statements correctly describes how a feature of this graph is related to the density of the fluid?
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?
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?
A fluid flows through the two sections of cylindrical pipe shown in the figure. The narrow section of the pipe has radius \( R \) and the wide section has radius \( 2R \). What is the ratio of the fluid’s speed in the wide section of pipe to its speed in the narrow section of pipe, \( \dfrac{v_{\text{wide}}}{v_{\text{narrow}}} \)?
The experimental diving rig is lowered from rest at the ocean’s surface and reaches a maximum depth of \(80\) \(\text{m}\). Initially it accelerates downward at a rate of \(0.10\) \(\text{m/s}^2\) until it reaches a speed of \(2.0\) \(\text{m/s}\), which then remains constant. During the descent, the pressure inside the bell remains constant at \(1\) atmosphere. The top of the bell has a cross-sectional area \(A = 9.0\) \(\text{m}^2\). The density of seawater is \(1025\) \(\text{kg/m}^3\).
Water flows from point \( A \) to points \( D \) and \( E \) as shown. Some of the flow parameters are known, as shown in the table. Determine the unknown parameters. Note the diagram above does not show the relative diameters of each section of the pipe.
| Section | Diameter | Flow Rate | Velocity |
|---|---|---|---|
| \( \text{AB} \) | \( 300 \) \( \text{mm} \) | \(\textbf{?}\) | \(\textbf{?}\) |
| \( \text{BC} \) | \( 600 \) \( \text{mm} \) | \(\textbf{?}\) | \( 1.2 \) \( \text{m/s} \) |
| \( \text{CD} \) | \(\textbf{?}\) | \( Q_{CD} = 2Q_{CE} \) \( \text{m}^3/\text{s} \) | \( 1.4 \) \( \text{m/s} \) |
| \( \text{CE} \) | \( 150 \) \( \text{mm} \) | \( Q_{CE} = 0.5Q_{CD} \) \( \text{m}^3/\text{s} \) | \(\textbf{?}\) |
Nancy is using a turkey baster (a kitchen tool with a rubber bulb on one end and a tube on the other) to collect juices from a roasting turkey. When she squeezes and then releases the rubber bulb, it creates suction with a pressure of \( 99{,}800 \) \( \text{Pa} \). This suction causes the turkey juice to rise \( 9 \) \( \text{cm} \) up the tube. Based on this information, what is the density of the turkey juice?
Why do you float higher in salt water than in fresh water?
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?
A solid titanium sphere of radius \( 0.35 \) \( \text{m} \) has a density \( 4500 \) \( \text{kg/m}^3 \). It is held suspended completely underwater by a cable. What is the tension in the cable?
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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] |
Metric Prefixes
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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