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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."
The drawing above shows a spherical reservoir that contains \( 455,000 \) \( \text{kg} \) of water when full. The reservoir is vented to the atmosphere at the top. Assuming the reservoir is full and the diameter of the reservoir is much larger than any of the pipes on the ground.
In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
A spherical balloon of mass \( 226 \) \( \text{kg} \) is filled with helium gas until its volume is \( 325 \) \( \text{m}^3 \). Assume the density of air is \( 1.29 \) \( \text{kg/m}^3 \) and the density of helium is \( 0.179 \) \( \text{kg/m}^3 \).
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?
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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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