Cheat Sheet for exam 1 - Winter Semester with Prof. Guenther PDF

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Winter Semester with Prof. Guenther...

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Numerical Constants and Conversions: −15 −12 −9 −6 −3 −2 | pico 10 | nano 10 |micro(µ) 10 | milli 10 |centi 10 femto 10 3 6 9 kilo 10 |Mega 10 | Giga 10 g = 9.80 m/s2 |1 in = 2.54 cm |1m = 39.4in| 12 in = 1 ft| 1 m = 3.281 ft| 1 mi = 1.609 km| 1 in = 2.54 cm |1m = 39.4in 12 in = 1 ft| 1 m = 3.281 ft| 1 mi = 5280 ft| 3.16 × 107 s = 1 year| 1m=0.001km| 100 cm= 1000 mm Vectors and 2D Motion Vectors have direction + magnitude whereas scalars just have magnitude Tip to Tail (vector) + (vector) = (vector) Subtraction: (vector) + (-1)(vector) = (vector) (scalar) * (vector) = (vector) E = 2D If you need to find C it is C= the square root of Cx^2+Cy^2 Cx=Ax+Bx & Cy=Ay+By Magnitude is never negative DIFFERENT GRAPHS REPRESENT DIFFERENT THINGS AND THE SLOPE ON THOSE GRAPHS MEAN DIFFERENT THINGS AS WELL V vs. T graphs The area under the graph gives the displacement of the object. The slope of the line is the acceleration A vs T graphs The area under the graph gives the total velocity of the object. The slope of the line is the velocity Note slope changes = changing velocity=non-Zero accelerations= curve line

Use Program ACCELERA for equations shown above still show all work

PM: If free-fall, acceleration 9.8 m/s2 downward AND IF choose y-axis vertical (+y up typically) ax = 0 m/s2 ,ay = -9.8m/s2 Average equations How to solve 1&2 D problems: Final Answer Equation ∆x ∆t Draw/Image – lay out coordinate system| Write down what ∆x m t f −t 0=s m a´ avg= x f − x0 = you know and what you want to find. |Pick the appropriate 2 ∆t s s formula or formulas| Solve

v´ avg =

∆x ∆t

x f − x0 =

m s

t f −t 0=s

m s

FORCES � = � g- this is weight which is always pulling down on the FBD for this particular equation g= 9.8 m s^2 Newton’s 1st Law - In the absence of external forces, when viewed from an inertial reference frame, an object at rest will remain at rest and an object in motion continues in motion with a constant velocity Newton’s 2nd Law - Net Force is the product of Mass and Acceleration ���� = � � Newton’s 3rd Law - If two objects interact, the force that object one is exerting on object 2 is equal and opposite to that object two is exerting on object one �� ,� = −� �,�. 1N= 1kg*m/s^2 What to do to figure out the answers 1. Determine your system 2. Setup your coordinates 3. Identify all forces and draw a Free Body Diagram (FBD) 4. Write down Newton’s 2nd Law, ���� = ��, in component form 5. Solve Elevator equations N = mg if the elevator is at rest or moving at constant velocity N = mg + ma if the elevator has an upward acceleration N = mg - ma if the elevator has a downward acceleration Asking for gravitational Force its mg = F and g is positive 9.8 unless given a different value Asking for net force solve normally Asking for normal force use one of the elevator equations Acceleration in an elevator you do g+a and g= 9.8 Deceleration in an elevator you do g-a and g=9.8...