Phys1 - Motion
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Meter Conversions
1 gigameter (G) = 10
^{9}
m
1 megameter (M) = 10
^{6}
m
1 kilometer (km) = 10
^{3}
m
1 centimeter (cm) = 10
^{-2}
m
1 millimeter (mm) = 10
^{-3}
m
1 micrometer (µm) = 10
^{-6}
m
1 nanometer (nm) = 10
^{-9}
m
1 Ångström (Å) = 10
^{-10}
m
1 picometer (pm) = 10
^{-12}
m
1 femptometer (fm) = 10
^{-15}
m
Other Conversions
√2 = 1.4
√3 = 1.7
sin30 = 1/2
cos30 = √3/2 ~ .86
sin60 = √3/2
cos60 = 1/2
sin & cos 45 = √2/2 ~ .7
Milligram Conversions
1 mg = 10
^{-3}
g = 10
^{-6}
kg
Distance
a Scalar quantity that refers to how much ground an object has covered during its motion
will only equal 0 if you don’t move at all
Displacement
a Vector quantity that refers to how far out of place an object is from where it started, aka it’s the object's overall change in position
displacement is the quantity that can equal 0 when you end in the same place you started
(average) Acceleration
change in velocity divided by the elapsed time for that change
a = v
_{2}
- v
_{1}
/ Δt
aka change in velocity / elapsed time
Most basic translational motion equation:
d = vt
distance (meters) = velocity (m/s) * time (s)
What equation should be used if vf is missing?
d = v
_{o}
t + 1/2at
^{2}
What equation should be used if d is missing?
v
_{f}
= v
_{o}
+ at
What equation should be used if t is missing?
v
_{f}
^{2}
= v
_{o}
^{2}
+ 2ad
Equation to find the time at the top of a projectile motion problem?
t
_{top}
= v
_{oy}
/g
same thing as saying m/s * s
^{2}
/m
Equation to find the height at the top of a projectile motion problem?
h = v
_{oy}
^{2}
/ 2g
What equation can be used to determine an object’s displacement during FREE-FALL?
d (Δh) = 5t
^{2}
from d = v
_{o}
t + 1/2at
^{2}
Objects with a ____ contact area-to-mass ratio experience a more significant impact of wind resistance than their denser counterparts.
Objects with a LARGE surface area : mass ratio experience MORE wind resistance than their denser (therefore heavier) counterparts
Terminal Velocity
• vterm occurs when the drag Force upward is equal & opposite to the Force of gravity (F
_{g}
)
• the object with the greatest terminal velocity will be:
1. the DENSEST
2. have the LEAST amount of contact area
Projectile Motion
v
_{ox}
= v
_{o}
cosθ
v
_{oy}
= v
_{o}
sinθ
Range Equations (3)
r = v
_{o}
^{2}
sin2θ / g
r = 2v
_{ox}
v
_{oy}
/ g
r
_{45°}
=
4
h
_{max}
Which launch angle results in the maximum range & which results in the maximum height?
• a 45° launch angle corresponds to the maximum possible RANGE
• a 90° launch angle corresponds to the maximum possible HEIGHT
• 30° & 60° launch angles correspond to the same range
Card Set Information
Author:
mse263
ID:
285169
Filename:
Phys1 - Motion
Updated:
2014-10-07 23:56:09
Tags:
Physics
Folders:
Physics
Description:
Chapter 1
Show Answers:
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