Step | Formula Derivation | Reasoning |
---|---|---|
1 | a = \frac{F}{m} | Calculating horizontal acceleration (a) from thrust (F) and mass (m). |
2 | s = ut + \frac{1}{2}at^2 | Distance formula for motion with constant acceleration. |
3 | 2.10, \text{m} = \frac{1}{2}a t_{\text{table}}^2 | Distance covered on the table is 2.10 m, initial velocity (u) is 0. |
4 | t_{\text{table}} = \sqrt{\frac{2 \times 2.10, \text{m}}{a}} | Solving for the time (t_{\text{table}}) taken to travel across the table. |
5 | v_{x} = at_{\text{table}} | Calculating the horizontal velocity (v_{x}) at the edge of the table. |
6 | v_{x} = 4.2988, m/s | Plug in values and solve. |
Step | Formula Derivation | Reasoning |
---|---|---|
1 | y = \frac{1}{2}gt_{\text{fall}}^2 | Distance formula for free fall, where y is the fall height and g is the acceleration due to gravity. |
2 | t_{\text{fall}} = \sqrt{\frac{2y}{g}} | Solving for the time (t_{\text{fall}}) to fall 2.10 m. |
3 | t_{\text{fall}} = .654, s | Plug in values and solve. |
Step | Formula Derivation | Reasoning |
---|---|---|
1 | d_{\text{air}} = v_{x} t_{\text{fall}} | Horizontal distance (d_{\text{air}}) traveled while falling, using the horizontal velocity at the edge of the table and the fall time. |
2 | d_{\text{air}} =4.3 * .65 | Plug in values to get d_{\text{air}} = 2.81, m. Final Answer. |
Phy can also check your working. Just snap a picture!
A truck is traveling at 35 m/s when the driver realizes the truck as no breaks. He sees a ramp off the road, inclined at 20°, and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction).
A textbook is launched up with a speed of 20 m/s, at an angle of 36°, from a 12 m high roof.
A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 18.9 m/s at an angle of 52.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
You must split an apple resting on top of you friend’s head from a distance of 27 m. When you aim directly at the apple, the arrow is horizontal. At what angle should you aim the arrow to hit the apple if the arrow travels at a speed of 35 m/s?
Determine the force needed to push a 150 kg body up a smooth 30° incline with an acceleration of 6 m/s2.
2.81 meters from the base of the table
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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_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \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 |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
k | Spring constant, in \text{N/m} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
Variable | SI Unit |
---|---|
s (Displacement) | \text{meters (m)} |
v (Velocity) | \text{meters per second (m/s)} |
a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
t (Time) | \text{seconds (s)} |
m (Mass) | \text{kilograms (kg)} |
Variable | Derived SI Unit |
---|---|
F (Force) | \text{newtons (N)} |
E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
P (Power) | \text{watts (W)} |
p (Momentum) | \text{kilogram meters per second (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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