Hi everyone,

                      Today we did lab and before that Ms.Kozoriz discussed WORK  "which is defined as the integral of force over a distance of displacement." 

        Work=Force.Distance

         w=f.d

Its unit is joule. POWER "it is the time rate at which work is done"

           Power=Work done/Time

           p=w/t

Its unit is watt. Then we did lab on work and power. Also we got a chapter 10 study guide on Energy and Power. Lab worksheet on centripetal force and problem 4 and 5 were marked and were given back to us. Thats all we did today. Hope this post will help those who missed the class.

                                                        GOOD LUCK

Tuesday - October 26

In today's physics class, we went over Transparency 6-3 Worksheet (Uniform Circular Motion) , and both sides of the Centripetal Acceleration and Centripetal Force worksheet. Just a reminder, we had to hand in questions 4 and 5 of Centripetal Force worksheet for marks!

We also had to pick up one worksheet (Acceleration and Circular Motion) and two readings (Centripetal and Centrifugal Forces, and Motion Near the Earth's Surface). Please have the worksheet done for tomorrow's class and also read through both readings.

Also, if you still have your lab from Thursday, it is now overdue!

Enjoy the rest of your night!
-- Rainer

Applet for Circular Motion

Circular Motion Applet
Take a look at the direction of the force vector and the acceleration vector.
What is the relationship between the tangential velocity vector and the acceleration vector?
What is the relationship between the tangential velocity vector and the force vector?
In what direction do both the force & acceleration vectors point? Can you explain why this is so?

Centripetal Force

Today, we did the lab on Centripetal Force. We got into groups and tested how long a rubber stopper would go around if we put 5 washers and so on. We will do this lab again, since we needed more time and also because some people were away.

Also a reminder that the worksheet we got yesterday, Centripetal Acceleration is due by Monday.

Projectile Motion

Hi Dear Classmates,

                                      today Ms.Kozoriz returned our chapter 6 REINFORCEMENT. After that Ms.Kozoriz solved the problems 1-4 from chapter 6 REVIEW the solution is:

1. horizontal velocity vh=800m/s

horizontal distance dh=180m

time t=dh/vh

t=180m/800m/s

t=0.225s

vertical distance dv=1/2g.tsquare

dv=1/2.9.8m/secsquare.0.225s.0.225s

dv=0.25m

2.dv=8.0m

dh=10.0m

t=under root 2dv/g

t=under root 2.8.0m/9.8m/s

t=1.27s

vh=dh/t

vh=10m/1.27s

vh=7.8m/s

3.

    a.    vh=45m/s

theta=45 degrees

vh=cos45 degree.45m/s=31.8m/s

vv=sin45 degrees.45m/s=31.8m/s

t=2vv/g

t=2.31.8m/s/9.8m/secsquare

t=6.48s

   b.  dh=vh.t

dh=31.8m/s.6.48s

dh=206.06m

4.

    a.  v=8m/s

theta=30 degrees

vh=cos30 degrees.8m/s=6.928m/s

vv=sin30 degrees.8m/s=4m/s

t=2vv/g

t=2.4m/s/9.8m/secsquare=0.81s

   b.  dv=v2 square-v1 square/2g

dv=o-16m/s/2.-9.8

dv=-16m/s/-19.6

dv=0.82m

    c.  dh=vh.t

dh=6.928m/s.0.81s

dh=5.61m

                   the remaining 5 problems are related to our next topics so we'll do them in future.

We also solved the green book's 2-7 problems. The solutions are:

2.vh=8m/s

dv=78.4m

t=under root 2dv/g

t=4s

dh=vh.t

dh=8.4

dh=32m

3.

   a.    t=under root vv/g

t=under root 1.225/9.8

t=under root .25

t=.5 s

    b.  vh=dh/t

vh=0.40/.5

vh=0.800m/s

4. vh=2.8m/s

t=2.6s

dv=?

dh=vh.t

dh=2.8.2.6

dh=7.26m

dv=1/2gtsquare

dv=33.1m

5.

   a.  dv=1001m

vh=125km/hr=34.7m/s

t=under root 2 dv/g

t=14.3s

   b.dh=vh.t

dh=34.7.14.3

dh=496.21m

6.dv=61m

dh=23m

vh=3m/s

t=under root 2dv/g

t=3.52s

vh=dh/t

vh=6.51

7. vy=vyt+1/2gt square  since vy=o

t=under root 2y/g

t=under root 2.-0.32/-9.8

t=0.26

x=vxt=12.4m/s . 0.26s

=3.2m

          Then we revised the formulas for projectile quiz. We did quiz about projectile. Thats all we did today and TOMORROW WE'LL HAVE A TEST ON MOMENTUM and Projectile Motion so be prepared.

      

                        BEST OF LUCK

Hello everyone,

                        Today we started with chapter 6 study guide which is about motion in two dimension. The answers are 

VOCABULARY REVIEW

 1. centripetal force

 2. projectile

 3. uniform circular motion

 4. centripetal acceleration

 5. trajectory

                                         PROJECTILE  MOTION

1. To an observer at position A the ball would appear moving straight up and straight down.

2. To an observer at position B the ball would appear moving in straight line.

3. To an observer at position C the ball would appear as an arc (parabola).

4. Throughout its flight,a projectile is constantly being accelerated toward the ground. Vertical vector shrinks until it reaches the top of its flight and then it increases as it approaches the ground. The horizontal velocity vector remains constant throughout its flight. Vertical vector is under the influence of gravity and horizontal vector has no net external force acting on it.

5. Neither rock lands first. Both rocks hit the ground at same time. This is occuring because both are under the same external force that is gravity(g).

6. Time t = ?

     d =8500m

      g=9.8m/s

      t=  2d/g

      t=2.8500m/9.8m/s

      t=1734.69

      t=41.64s      answer 

7. Vertical velocity v=?

     a=vf-vi/t

     v=a.t

     v=9.8m/s(42s)

     v=410m/s

8. dh=?

    Horizontal velocity v=483km/hr=134m/s 

    d=v.t

    d=483m/s.42s

    d=5600m

       Ms.Kzoriz wrote the answers for transparency sheet 6-1 and 2 on whiteboard. The answers for 6-1 are:

1. The magnitude of the ball's velocity vector is smallest at its maximum height.

2. Equal in magnitude but opposite in direction.

3. Equal in magnitude and in direction.

4. The line for horizontal position is a straight line and for vertical position its an arc.

        The answers for 6-2 are:

1. Same vertical vector at each point.

2. Both balls fall at same acceleration.

3. Horizontal motion of red ball does not affect its vertical motion.

4. Vertical interval is greater near the bottom of diagram. Since the time interval between each two picture is same, vertical vector of ball must have increased.

5. Horizontal interval between each two pictures of ball are identical. Since the intervals represent identical time periods. Horizontal vectors of the ball must be constant.

            Ms.Kozoriz talked about upwardly launched projectiles and we started to solve the problems on page 11 PROJECTILE MOTION 2. While doing problems we enjoyed a song Harry the Human Cannonball. Thats all we did in class today. One other important thing is ourshort  momentum test will be on Tuesday not on Monday.

      Its my first blog so if you guys found any mistakes sorry about that.

Correction to yesterday's post

Horizontal velocity is independent of any force acting in the x direction (not time like you stated in your post). Vertical velocity is dependent on the acceleration due to gravity acting on the object.

Physics12Fall2010: Scribe List

Projectile motion,this is a non straight motion under gravity (g).

the horizontal component of velocity is constant i.e its independent of time,but the vertical component of the velocity (Vy)depends on time. it's 0 at the peak

october 12 scribe

October 12 2010

hi! today, we went over the Transparency worksheet.

Then miss K review a little about momentum (elastic, explosion, inelastic).

After that we have a lab with money :) and hand in after class.

Sorry!!!!!!! i don't know how to put answers or picture up. If you needed you can ask miss K

Oh yeah! we picked up study guide "Motion in Two Dimensions" and a graph. Take a look we'll go over it tomorrow...

October 7

On Thursday, we went over the answers to the Chapter 9 Review worksheet.

Here are the answers:

1: 1600 N

2: 0.0013 s

3: 6700 N

4: 0.05 m/s [W]

5: -8.8 m/s

6: 81 m/s

7: 5.7 x 10^-2 m/s

8: 6 kg m/s

We then went on to finish the momentum unit by looking at momentum in two dimensions. These kind of problems involve finding the missing side of a triangle using a combination of the Law of Conservation of Momentum and trig ratios.

For homework, we got the Chapter 7: Momentum and Transparency 9-3 worksheets do.

Momentum Notes

Momentum

View more documents from ekozoriz.

Pool Table Applet

Pool Table Applet

Janna's Scribe for October 4th, 2010

Heyy everyone,

to start off class we watched a video that explored the concepts on impulse and momentum.

After watching the short video we went over the answers to the Chapter 9 Study Guide. The answers are as follows:

Section 9.1: IMPULSE and MOMENTUM

1. B

2. B

3. B

4. A

5. D

6. B

7. D

8. B

Next we went over the IMPULSE AND MOMENTUM sheet

1. WILL BE POSTED LATER

2.

a) m=Fg/g

=900N/9.8N/kg

=92kg

F∆t = m∆v/ ∆t

= (92kg)(2.0m/s)/ 0.70s

= 260N

b) p=m∆v

= (92kg)(2.0m/s)

=184N

260N - 184N =76N

3.

a) Ft =mv

=(51kg)(20.0m/s)

= 1020kg . m/s

b) F = mv/t seatbelt

= 1020kg.m/s/ 0.12s

= 8500N

F = mv/t noseatbelt
= 1020kg.m/s/ 0.008s
= 128000N

ALWAYS WEAR A SEATBLET WHEN IN A VEHICLE !!

Don't need to do #4

Next we watched three short clips regarding the momentum of a ball, recoil and the conservation of momemtum.

KEY CONCEPTS:

Remember momentum before a collision is equal to the momentum after a collision occurs.

For every action there is an equal and opposite reaction.

Homework:

Transparency 9-1 and 9-2 are due on Wednesday in class.

Have an awesome afternoon off !!! :)

Friday - October 1

Friday's physics class was basically a chance for those who missed the momentum lab on Thursday to do and finish the lab.

For those of us who were in class on Thursday, our task was to finish and hand in the lab.

So, reminder to those who didn't finish or forgot to hand in their lab, it is now over due!!



Next thing on our to do list was to check the answers from Wednesday's questions from the duck book.

Basically, to find the momentum from a given force vs. time graph, we find the area under the given line.

If the line creates a triangle, its area can be found using the equation area = (1/5)(base)(height).

If the line creates a rectangle, its area can be found using the equation area = (base)(height).

If the line creates an irregular shape, simply cut up the shape into pieces of triangles and/or rectangles and find each area using the equations from above. (Don't forget to add up all the areas in the end).


The rest of the class should have been spent on doing questions 1 - 4 of the Impulse and Momentum worksheet.

If you did not get a chance to finish it in class, please do it for homework!



Enjoy the rest of your weekend!!!
-- RAINER