LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(
Transcript of LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(
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UNIT 2 TEST REVIEW
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MY ADVICE
TAKE YOUR TIME!
Read Questions Carefully
Read all possible answer choices before choosing
Work Pad for Calculations-
if it is not working type your answer as a comment
Use degree mode!!!
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I AM REALLY NICE….BUT
THE PRACTICE TEST AND TEST NUMBER/ORDER DO NOT MATCH
MAKE SURE TO COMPLETE PRACTICE TEST & CHECK THE KEY
BEFORE YOU TAKE UNIT 2 TEST!
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OFFICE HOURSTUESDAY 9AM – 10AM
ORCALL/TEXT ANY QUESTIONS YOU
HAVE BEFORE TEST!
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• Given: vi = 27.0m/s teammate=82.0m away
• Make sure you are in degree mode for all calcs!
• draw pic
23.
• Find horizontal and vertical velocity components
•
•
• = 21.7m/s = 16.1m/s
• Along the curve the change in V for vertical motion can be found by:
•
• = -16.1 – (16.1) • 𝑣 negative because downward motion on second half of curve
• = -32.2 m/s
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• Find the time the ball is in the air
• solve for
•
•.
.
• 3.28 sec
• Find the horizontal displacement
•
• 21.7 (3.28)
• =71.2 m
• Teammate is 80m away. How far must he walk to catch ball?
•
• 8.8 m
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24.
37.8 North
29.5 North
? South – remember headwind is direct in opposite direction
• Velocity measured at ground is the resultant velocity
• falcon velocity + wind velocity = 29.5 m/s North
• = 37.8 + (x) = 29.5
• x = 29.5-37.8
• = -8.30 m/s
• = 8.30 m/s due south
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25.
• How do we find the westward motion?
• Subtract the east from the west.
• 1.2 x 103 -1.5
• 1198.5
• Label what we know
• Find displacement using Triangle Method
1,198.5 m
2.5 x102 m
• d =
•
•
• =1224.2
• =1.2 x 103 m North of West
displacement m
Add to
Notes
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26.
• Given info:
• F = 148N
• mass = 544kg
• acceleration = ?
• We use
• F= ma
• Solve for a
• a =
• = 148/544
• = .27205
• a = .272 m/s2
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• Given info:
• 1 brick = 8.0N
• .31
• 4 bricks = 32N
•
• Solve for
• =
• Is 8N a mass?
• It is actually the
• =
• = .31 x 32
• = 9.92 N
• = 9.9 N to begin motion
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𝑣 = 12.5𝑚/𝑠
• Given info:
•
• - 9.81m/s2
• (vf)2 = (vi)2 + 2aΔx• Solve for Δx
•
• =
• = -156.25/-19.62
• = 7.9638
• = 7.96m
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236,000,000,000,000,000,000 m
1,000 m
1 km= 236,000,000,000,000,000 km
236,000,000,000,000,000 km
= 2.36 x 1017 km7.
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Interpreting Graphs
What does this graph show us?
It shows that the object is accelerating as the velocity increases.
Instantaneous Velocity is the found by the slope of the tangent line
10.
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DON’T FORGETVECTOR ADDITION
TRIANGLE METHOD
• Tail of one vector is placed at the head of the other.
• The resultant is the vector drawn from the tail of the first to the head of the last vector.
12.
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13.
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DON’T FORGET
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d1
d2
d3
d4
d5
d6
Which displacement vectors shown have vertical components that are equal?
d4 and d6
vertical component = x value
19.
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20.MOTION OF BALL APPEARS
HOW??
To Player – It looks like it drops directly downward because
moving at same velocity (Only true for small velocities)
To Crowd – appears to drop downward and backwards from
players position
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BRING ON THE AIR RESISTANCE
21.
Wind/Air Resistance
Force of Gravity
F=mam = 80kg
Fg = 784.8 N
Fg = 780 N
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SCIENTIFIC NOTATION REVIEW
Operation Property Example
addition and subtraction power of 10 must have same exponent – adjust whole number then add*if number is greater than 9 must adjust back into correct notation
3 x 104
+ 8 x 104
11 x 104
= 1.1 x 105
multiplication am x bn = a x b m+n (2x104) x (8x109) = 16 x 1013
=1.6 x 1014
division am ÷ bn = a x b m-n 5.6 x 109 / 4.5 x 105
=1.2 x 104
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• (2.99 + 2.20 x 103) x (5.26 x 10-2)
• (2.99+2200)
• (2.20299 x 103) x (5.26 x 10-2)
• 11.5877 x 101
• 1.16 x 102
22.