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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = v_i + at \) | Use the first equation of motion to relate the initial velocity, final velocity, and time. |
| 2 | \(0 \, \text{m/s} = 5.0 \, \text{m/s} – 9.8 \, \text{m/s}^2 \cdot t \) | At maximum height, the final velocity \(v_f\) is zero, the initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), and the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\) (due to gravity). |
| 3 | \( t = \frac{5.0 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 0.51 \, \text{s} \) | Calculate the time interval for the rock to reach its maximum height. This is only the time to reach the maximum height, so the total time to fall back to the original location is double this time. |
| 4 | \( t_{\text{total}} = 2 \cdot 0.51 \, \text{s} \approx 1.02 \, \text{s} \) | Multiply by 2 to get the total time interval for the rock to return to its original location. |
| 5 | \(\boxed{1.02 \, \text{s}}\) | Final answer for part (a) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = v_i + at \) | Use the first equation of motion to relate the initial velocity, final velocity, and time. |
| 2 | \(v_f = 5.0 \, \text{m/s} – 9.8 \, \text{m/s}^2 \cdot 1.02 \, \text{s} \) | The initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\) (due to gravity), and the total time \(t\) is \(1.02 \, \text{s}\). |
| 3 | \(v_f \approx 5.0 \, \text{m/s} – 10.0 \, \text{m/s} = -5.0 \, \text{m/s} \) | Calculate the final velocity when the rock returns to the same height. Note the negative sign indicates the direction is downward. |
| 4 | \(\boxed{-5.0 \, \text{m/s}}\) | Final answer for part (b). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(a = -g\) | Acceleration due to gravity is always acting downward. |
| 2 | \(a = -9.8 \, \text{m/s}^2\) | Even at maximum height, the acceleration due to gravity remains \(-9.8 \, \text{m/s}^2\). |
| 3 | \(\boxed{-9.8 \, \text{m/s}^2}\) | Final answer for part (c). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = 0 \, \text{m/s}\) | At maximum height, the velocity of the rock is zero as it changes direction. |
| 2 | \(\boxed{0 \, \text{m/s}}\) | Final answer for part (d) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f^2 = v_i^2 + 2a\Delta x\) | Use the third equation of motion to relate initial velocity, final velocity, acceleration, and displacement. |
| 2 | \(0 = (5.0 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)\Delta x\) | At maximum height, the final velocity \(v_f\) is zero. The initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), and acceleration \(a\) is \(-9.8 \, \text{m/s}^2\). |
| 3 | \(0 = 25 – 19.6 \Delta x \) | Simplify the equation. |
| 4 | \( 19.6 \Delta x = 25 \) | Rearrange the equation to solve for height \( \Delta x \). |
| 5 | \( \Delta x = \frac{25}{19.6} \approx 1.28 \, \text{m} \) | Calculate the maximum height using the final value obtained from the rearranged equation. |
| 6 | \( \boxed{1.28 \, \text{m}} \) | Final answer for part (e). |
Just ask: "Help me solve this problem."
An airplane accelerates down a runway at \( 10 \, \text{m/s}^2 \). It reaches a final velocity of \( 200 \, \text{m/s} \) until it finally lifts off the ground. Determine the distance traveled before takeoff.
A red car, initially at rest, travels east with an acceleration of \( 3.5 \, \text{m/s}^2 \). At the same time as the red car starts to move, a blue car is traveling west at \( 15 \, \text{m/s} \) and accelerating at \( 1.2 \, \text{m/s}^2 \). If they are \( 600 \, \text{m} \) apart the moment the red car starts to move and they are traveling towards each other, where and when will they meet?
A coin is dropped from a hot air-balloon that is 250 m above the ground rising at 11 m/s upwards. For the coin, assume up is positive and find the following:
A car travels at \( 20 \, \text{m/s} \) for 5 minutes and then travels another 2 km at \( 40 \, \text{m/s} \). What is the total distance traveled and time of travel for the car?
A) \( 1.02 \, \text{s} \)
B) \( -5 \, \text{m/s} \)
C) \( -9.8 \, \text{m/s}^2 \)
D) \( 0 \, \text{m/s} \)
E) \( 1.28 \, \text{m} \)
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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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