Part A – Buoyant force
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) 1 | \(B = W_{\text{air}} – W_{\text{water}}\) | The buoyant force \(B\) equals the difference between the weight of the object measured in air and the apparent weight when submerged. |
| (a) 2 | \(B = 17.8\,N – 16.2\,N = 1.6\,N\) | Substitute the given readings to calculate the buoyant force. |
| (a) 3 | \(\boxed{B = 1.6\,N}\) | This is the final buoyant force acting on the object in water. |
Part B – Volume of the object
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (b) 1 | \(B = \rho_{w} g V\) | According to Archimedes’ principle, the buoyant force is equal to the weight of the displaced water where \(\rho_{w}\) is the density of water, \(g\) is gravitational acceleration, and \(V\) is the volume displaced. |
| (b) 2 | \(V = \frac{B}{\rho_{w} g}\) | Rearrange the formula to solve for the volume of the object. |
| (b) 3 | \(V = \frac{1.6}{1000 \times 9.8}\) | Substitute \(B = 1.6\,N\), \(\rho_{w} = 1000\,kg/m^3\), and \(g = 9.8\,m/s^2\). |
| (b) 4 | \(V \approx 1.63 \times 10^{-4}\,m^3\) | Compute the division \(1.6/(9800)\) to obtain the object’s volume. |
| (b) 5 | \(\boxed{V \approx 1.63 \times 10^{-4}\,m^3}\) | This is the final volume of the object. |
Part C – Density of the object
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (c) 1 | \(W = m g\) | The weight of the object in air is the product of its mass \(m\) and gravitational acceleration \(g\). |
| (c) 2 | \(m = \frac{W}{g} = \frac{17.8}{9.8}\) | Solve for the mass by rearranging the weight formula using \(W = 17.8\,N\) and \(g = 9.8\,m/s^2\). |
| (c) 3 | \(m \approx 1.82\,kg\) | Performing the division gives the mass of the object. |
| (c) 4 | \(\rho = \frac{m}{V}\) | Density is defined as mass divided by volume. |
| (c) 5 | \(\rho = \frac{1.82}{1.63 \times 10^{-4}}\) | Substitute \(m \approx 1.82\,kg\) and \(V \approx 1.63 \times 10^{-4}\,m^3\) into the density formula. |
| (c) 6 | \(\rho \approx 1.12 \times 10^4\,kg/m^3\) | The division yields the density of the object. |
| (c) 7 | \(\boxed{\rho \approx 1.12 \times 10^4\,kg/m^3}\) | This is the final density of the object. |
Part D – Absolute pressure when object is removed
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (d) 1 | \(p = p_{\text{atm}} + \rho_{w} g h\) | The absolute pressure at the bottom of a water column is given by the sum of atmospheric pressure \(p_{\text{atm}}\) and the hydrostatic pressure \(\rho_{w} g h\), where \(h\) is the water depth. |
| (d) 2 | Removing the object | Upon removal of the object, the water can now fill the space the ball occupied. This reduces the overall water depth \(h\). |
| (d) 4 | \(\boxed{p \text{ decreases}}\) | Thus, the hydrostatic pressure (\(\rho gh\)) decreases, due to the decrease in height of the water. Since hydrostatic pressure drops so will the absolute pressure, as given by the equation in (d) 1 |
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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?
Suppose we wish to make a neutrally buoyant hollow sphere out of titanium (\(\rho = 4500 \text{kg/m}^3\)). If the sphere has an outer radius of \( 1.5 \) \( \text{m} \), what must be its inner radius?
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?
The radius of the aorta is about \( 1 \) \( \text{cm} \) and the blood flowing through it has a speed of about \( 30 \) \( \frac{\text{cm}}{\text{s}} \). Calculate the average speed of the blood in the capillaries given the total cross section of all the capillaries is about \( 2000 \) \( \text{cm}^2 \).
A cube of side length \( s \) rests on the bottom surface of a container of fluid. The fluid is at a height \( y \) above the bottom of the tank. The fluid has density \( \rho \) and the atmospheric pressure is \( P_{\text{atm}} \).
Which of the following expressions is equal to the absolute pressure exerted by the fluid on the top surface of the cube?
Wanda watches the fish in her fish tank and notices that the angelfish like to feed at the water’s surface, while the catfish feed \( 0.300 \) \( \text{m} \) below at the bottom of the tank. If the average density of the water in the tank is \( 1000\) \( \text{kg/m}^3 \), what is the absolute pressure on the catfish?
How large must a heating duct be if air moving \( 3 \ \frac{\text{m}}{\text{s}} \) along it can replenish the air in a room of \( 300 \ \text{m}^3 \) volume every \( 15 \) minutes? Assume the air’s density remains constant.
Water of density \( \rho \) flows through the section of circular pipe shown in the figure. At Point A, where the diameter of the pipe is \( D \), the water has a pressure \( P_0 \) and velocity \( v_0 \). Point B is located a vertical distance \( H \) above Point A in a section of the pipe that has diameter \( 2D \).
Which of the following expressions is equal to the pressure of the water at Point B?
The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is \( 2.5 \) \( \text{m} \), and the puncture is \( 1 \) \( \text{m} \) above the ground level, what is the efflux speed of the water?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
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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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