ll5 u2 review - physics€¦ · 0< $'9,&(7$.(

21
UNIT 2 TEST REVIEW

Upload: others

Post on 07-Oct-2020

0 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

UNIT 2 TEST REVIEW

Page 2: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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!!!

Page 3: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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!

Page 4: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

OFFICE HOURSTUESDAY 9AM – 10AM

ORCALL/TEXT ANY QUESTIONS YOU

HAVE BEFORE TEST!

Page 5: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

• 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

Page 6: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

• 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

Page 7: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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

Page 8: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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

Page 9: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

26.

• Given info:

• F = 148N

• mass = 544kg

• acceleration = ?

• We use

• F= ma

• Solve for a

• a =

• = 148/544

• = .27205

• a = .272 m/s2

Page 10: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

• 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

Page 11: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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

Page 12: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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.

Page 13: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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.

Page 14: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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.

Page 15: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

13.

Page 16: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

DON’T FORGET

Page 17: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

d1

d2

d3

d4

d5

d6

Which displacement vectors shown have vertical components that are equal?

d4 and d6

vertical component = x value

19.

Page 18: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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

Page 19: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

BRING ON THE AIR RESISTANCE

21.

Wind/Air Resistance

Force of Gravity

F=mam = 80kg

Fg = 784.8 N

Fg = 780 N

Page 20: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

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

Page 21: LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

• (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.