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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v = v_0 + at \) | Use the kinematic equation for vertical motion. Here, \(v\) is the final velocity at the max height (which is 0), \(v_0\) is the initial velocity, \(a\) is the acceleration (gravity, acting downward), and \(t\) is the time. |
| 2 | \( 0 = v_0 – g \times 0.586 \) | Set the final velocity \(v\) at the maximum height to 0, and solve for \(v_0\). The acceleration due to gravity \(g\) is \(9.8 \, \text{m/s}^2\). |
| 3 | \( v_0 = g \times 0.586 \) | Rearrange to solve for \(v_0\) |
| 4 | \( v_0 = 9.8 \times 0.586 \) | Substitute the values of \(g\) and \(t\) |
| 5 | \( v_0 = 5.74 \, \text{m/s} \) | Final answer for the initial speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v^2 = v_0^2 + 2a \Delta x \) | Use the kinematic equation to relate velocity, acceleration, and displacement. Here, \(v\) is the final velocity at the max height (0), \(v_0\) is the initial velocity, \(a\) is the acceleration (gravity), and \(\Delta x\) is the displacement. |
| 2 | \( 0 = (5.74)^2 – 2 \cdot 9.8 \cdot \Delta x \) | Substitute \(v = 0\), \(v_0 = 5.74 \, \text{m/s}\), and \(a = 9.8 \, \text{m/s}^2\) |
| 3 | \( 2 \cdot 9.8 \cdot \Delta x = (5.74)^2 \) | Rearrange to solve for \(\Delta x\). |
| 4 | \( \Delta x = \frac{(5.74)^2}{2 \cdot 9.8} \) | Solve for \(\Delta x\). |
| 5 | \( \Delta x = 1.68 \, \text{m} \) | Compute the displacement \(\Delta x\). |
| 6 | \(\text{Max height} = 3.25 + 1.68 = 4.93 \, \text{m} \) | Since the ice cube was initially 3.25 m above the ground, add this to \(\Delta x\) to get the maximum height. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\Delta y = v_0 t + \frac{1}{2} a t^2\) | Use the kinematic equation for vertical displacement, where \(\Delta y\) is the change in height, \(v_0\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(t\) is the time. |
| 2 | \(\Delta y = 0\) at maximum height, then use \(y = 4.93 \, \text{m}\) | The maximum height the ice cube reaches is previously calculated as 4.93 m. |
| 3 | \( -4.93 \, \text{m} = – \frac{1}{2} \cdot 9.8 \cdot t^2\) | Solve for the time taken to fall from the maximum height to the ground. Note the displacement (\(\Delta y\)) is negative when falling down. |
| 4 | \(t = \sqrt{\frac{2 \cdot 4.93}{9.8}}\) | Substitute the values and solve for \(t\). |
| 5 | \(t = \sqrt{\frac{9.86}{9.8}} = \sqrt{1.01} \approx 1.005 \, \text{s}\) | Calculate the square root to find the fall time. |
| 6 | Total time = \(0.586 \, \text{s} (up) + 1.005\, \text{s} (down) \approx 1.59 \, \text{s}\) | Add the time going up (0.586 s) to the time coming down (1.005 s) to get the total time. |
| 7 | \( t_{\text{total}} = 1.59 \, \text{s} \) | Final time taken for the ice cube to reach the ground. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y^2 = v_{max}^2 + 2 a \Delta y\) | Use the kinematic equation relating velocity, acceleration, and displacement, where \(v_y\) is the final velocity, \(v_{max}\) is the velocity at maximum height (0 m/s), \(a\) is the acceleration due to gravity, and \(\Delta y\) is the displacement (4.93 m). |
| 2 | \(v_y^2 = 0 + 2 \cdot 9.8 \cdot 4.93\) | Substitute the values into the equation. |
| 3 | \( v_y^2 = 96.508 \) | Calculate the right side of the equation. |
| 4 | \( v_y = \sqrt{96.508}\) | Take the square root to solve for \(v_y\). |
| 5 | \( v_y \approx 9.82 \, \text{m/s}\) | Final speed of the ice cube when it reaches the ground. |
| 6 | \( v_{\text{final}} \approx 9.82 \, \text{m/s} \) | The boxed final answer for the speed of the ice cube when it hits the ground. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y^2 = v_{max}^2 + 2 a \Delta y\) | Use the kinematic equation relating velocity, acceleration, and displacement, where \(v_y\) is the final velocity (7.00 m/s), \(v_{max}\) is the velocity at maximum height (0 m/s), \(a\) is the acceleration due to gravity, and \(\Delta y\) is the displacement from the maximum height. |
| 2 | \( (7.00)^2 = 0 + 2 \cdot 9.8 \cdot \Delta y \) | Substitute the given speed and constants. |
| 3 | \( 49 = 19.6 \cdot \Delta y\) | Solve for \(\Delta y\) by isolating it on one side of the equation. |
| 4 | \( \Delta y = \frac{49}{19.6}\) | Rearrange to solve for \(\Delta y\). |
| 5 | \(\Delta y = 2.5 \, \text{m}\) | Final height change from the maximum height when the ice cube reaches a speed of 7.00 m/s downward. |
| 6 | \( \text{Height above ground} = 4.93 – 2.5 \) | Subtract the downwards displacement from the maximum height to get the current height above the ground. |
| 7 | \( \text{Height above ground} = 2.43 \, \text{m}\) | Final height of the ice cube above the ground when traveling at 7.00 m/s downward. |
| 8 | \( \text{Height} = 2.43 \, \text{m} \) | Boxed final answer. |
Just ask: "Help me solve this problem."
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The motion of a particle is described in the velocity vs. time graph shown above. Over the nine-second interval shown, we can say that the speed of the particle…
A block is projected up a ramp with an initial speed \( v_0 \). It travels along the surface of the ramp with constant acceleration \( a \). Take the positive direction of motion to be up the ramp. If the acceleration vector points opposite the initial velocity vector, which of the following MUST be true?
Above is the graph of the velocity vs. time of a duck flying due south for the winter. At what point might the duck begin reversing directions?
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
Note answers may vary by \( \pm 0.2 \).
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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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