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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 |
Just ask: "Help me solve this problem."
A fountain with an opening of radius \( 0.015 \) \( \text{m} \) shoots a stream of water vertically from ground level at \( 6.0 \) \( \text{m/s} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
Johnny the auto mechanic is raising a \( 1200 \) \( \text{kg} \) car on her hydraulic lift so that she can work underneath. If the area of the input piston is \( 12 \) \( \text{cm}^2 \), 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?
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 \(1025\) \(\text{kg/m}^3\).
The radius of the left piston is \( 0.12 \) \( \text{m} \) and the radius of the right piston is \( 0.65 \) \( \text{m} \). If \( f \) were raised by \( 14 \) \( \text{N} \), how much would \( F \) need to be increased to maintain equilibrium?
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?
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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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