0 attempts

0% avg

UBQ Credits

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | [katex] F = 23 \, \text{N} [/katex] | The force exerted by the person on the crank handle is given as 23 N. |

2 | [katex] r = 0.25 \, \text{m} [/katex] | The radius of the circular path described by the handle is given as 0.25 m. |

3 | [katex] T = 1.7 \, \text{s} [/katex] | The time period for one complete turn of the handle is given as 1.7 seconds. |

4 | [katex] \text{Circumference} = 2\pi r [/katex] | Calculate the circumference of the circular path which the handle moves in one complete revolution. |

5 | [katex] \text{Circumference} = 2\pi \times 0.25 \, \text{m} = 0.5\pi \, \text{m} [/katex] | Substitute the value of [katex] r [/katex] to find the circumference of the path. |

6 | [katex] \text{Distance} = \text{Circumference} = 0.5\pi \, \text{m} [/katex] | The distance moved by the handle in one rotation equals the circumference of the circle. |

7 | [katex] \text{Work done} = F \times \text{Distance} [/katex] | The work done in one revolution can be calculated using the force and the distance moved in the direction of the force. |

8 | [katex] \text{Work done} = 23 \, \text{N} \times 0.5\pi \, \text{m} = 11.5\pi \, \text{J} [/katex] | Substitute the values of [katex] F [/katex] and [katex] \text{Distance} [/katex] to calculate the work done in one revolution. |

9 | [katex] \text{Power} = \frac{\text{Work done}}{T} [/katex] | Power is defined as the rate of doing work, thus power can be calculated as work done per unit time. |

10 | [katex] \text{Power} = \frac{11.5\pi \, \text{J}}{1.7 \, \text{s}} \approx 21.2 \, \text{W} [/katex] | Substituting the values of work done and time period to calculate the average power. |

Just ask: "Help me solve this problem."

- Statistics

Beginner

Mathematical

FRQ

If a small motor does 520 J of work to move a toy car 260 meters in a time of 37 seconds.

- Energy

Intermediate

Mathematical

GQ

A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 10^{3} N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

- Energy, Springs

Advanced

Mathematical

FRQ

A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.

- Energy, Linear Forces

Intermediate

Mathematical

GQ

An object with a mass m = 80 g is attached to a spring with a force constant k = 25 N/m. The spring is stretched 52.0 cm and released from rest. If it is oscillating on a horizontal frictionless surface, determine the velocity of the mass when it is halfway to the equilibrium position.

- Energy

Advanced

Mathematical

MCQ

A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.

What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?

- Energy

21.2 Watts

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

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

Conversion Example

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] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages and Images
- Unlimited UBQ Credits
- 157% Better than GPT
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- All Smart Actions
- Mobile Snaps
- Focus Mode
- No Ads

A quick explanation

Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.

Submitting counts as 1 attempt.

Viewing answers or explanations count as a failed attempts.

Phy gives partial credit if needed

MCQs and GQs are are 1 point each. FRQs will state points for each part.

Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.

Understand you mistakes quicker.

Phy automatically provides feedback so you can improve your responses.

10 Free Credits To Get You Started