Hi everyone
here are the correct formulas from yesterday's class:


Escape Velocity/speed:
1.
Kinetic energy = Potential energy
Notation form: Ek = Ep

1/2mv² = Gm₁m₂

V=√2Gm/r
2. Determining g of a planet:

Fg=mg Fg=Gm₁m₂/r²

g=Gm/r²

3. Determining r, m, T:

Fc=m4π²r/T² Fg=Gm₁m₂/r²

4π²r/T²=Gm₂/r²
solve for m:
m= 4π²r³/GT²

solve for T: T²= 4π²r³/Gm

* period and radius is all you need to use this formula

4. Orbital Velocity:Fc=mv²/r Fg= Gm₁m₂/r²
v²=Gm₂/r

** also remember that force formulas have r² (squared)but energy formulas have only r radius.

April 30, 2010 12:02 AM

Orbits

Thursday topic:

-We had a little quiz on Chapter 14: Satellite Motion ( CONCEPT-DEVELOPMENT PAGE) and everybody had the hand-out.

-We also answered the (Gravitational Potential Energy Questions).
Answer:
1. -3.13*10^10 J
2. -2.36*10^9 J
3. +2.98*10^9 Jjavascript:void(0)
4. 2.89*10^9 J
5. -1.18*10^11 J



-Ms. K also gave us New Formula(s):

Escape Velocity
r = Square root 2GM/R


Determining g of a Planet
g = Gm/R^2


Determining R, m, T
T = 2(3.14) Square Root R^3/R^2


Orbital Velocity
v = Square Root Gm/R



-We also had a hand-out about:

( 12.2 The Falling Moon )- It talks about Isaac Newton`s history of how he got things and especially the gravity.

( 14.5 Escaped Speed) - It also talks more on satellite positioning and orbit of earth to sun and etc. Masses, Radius and the Escaped Speed of particular planets in our solar system.

( Mass of a Central Object and Radius of Motion of a Satellite )- It shows the solution in finding the speed of a satellite.
v = 2(3.14)R/T

for m,


m = 4(3.14)^2R^3/GT^2


- lastly Ms. K gave us an assignment REVIEW Chapter 8. (1 - 7).



And that's all folks..the next scribe is Paullin? Camille? or Nailuj89?..

Orbits

Summary:
-Last Tuesday we watch the about Conceptual Physics by Paul Hewitt.
-We also had an assignment ( Gravitational Potential Energy Questions 1 - 5 )
-Assignment also in what we watched and must be 7 key point(s).


Key Points:

- In order the satellite didn't fall on Earth must run 8 km/s
- When it runs less than 8 km/s, in will fall.
- and the time in moves more than 8 km/s, it go in outer space and produce an ellipse orbit.


And that pretty much we discussed last Tuesday.

Blog for April 26th

wow...i forgot to do this last night...

Today in class we:
-handed in the chapter assignment (the one that requires word answers on Katie's post)
-Corrected the Chapter 8 Study Guide
-Corrected the Universal Law of Gravitation Worksheet
-Received 2 new sheets: Defining Gravitational Potential Energy and Gravitational Potential Energy Questions

Here are the answers for the Universal Law of Gravitation worsheet:
1) 7.41*10^-12 N
2) Answers on question 2 vary depending on your weight
3) a - 2 fold
b - 1/4
c - 4 fold
d - 16 fold
4) 1.62*M m/s^2, M = mass of object

I don't have time to post the rest of the answers up for the chapter 8 study guide so i suggest you borrow the answers from someone from class if you haven't already received them. Or you can pray that I'll post the answers and solutions to the questions another time.

Scribe for today is Karen

OOPPPSSS!!!!

OK I FORGOT HAHA WHO'S THE NEXT SCRIBE?!?!?!

I CHOOSE.... Paullin!!! :P psst.. good luck! xD

Friday!

Hey!!!! Hope everyone's having a great weekend. So as you may have figured out I'm the scribe for Friday April 23rd. I believe the first thing we did, or one of the first things, was answer the questions that were on the board when the subs were in class. Apparently I forgot to write down the first set of questions, but here are the answers for the second set.

P. 172 8-13

8. 5.8x10^-10 N

9. 8.0x10^10 N

10. 6.48x10^-8 N

11.A) 491 N

B) 490N

12. 9.01x10^-31 kg

13. m1= 0.37 kg m2= 0.75 kg

Ms. K desided that the concept questions on page 163 are due on Monday... hand in assignment.. eekk.. hope everyone did it and has a way of getting a green book to do it, if not then here:

1.1 Earth is attracted to the sun by the force of gravity. Why doesn't Earth fall into the sun? Explain.

1.2 If Earth began to shrinkbut it's mass remained the same, predict what would happen to the value of "g" on shrinking Earth's surface.

1.3 Cavendish did his experiment using lead spheres. Suppose he had used equal masses of copper instead. Would his value of G be the same or different?

1.4 Critical thinking: In a space colony on the moon, an astronaut can pick up a rock with less effort than on earth. Is this also true when she throws the rock horizontally? If the rock drops on here toe, will it hurt more or less than on earth? Explain your answers.

Well thats it for now hope you all have a wonderful weekend, and YAY! My first scribe is done!!!!! haha.

BYE!!!

Universal Gravitation

This is for the Thursday Blog:

Just like I said, it's still the same, and everybody is busy doing their own stuff. But we have an assignment or work problem about:

Concept Review: 1-4 pages 163
Problems: 8-12 pages 172

Concept Review possible answer:
1. The earth doesn't fall because it is revolving and rotating at the same time around the sun.

2. I can't figure it out and it's kind of confusing.

3. Sorry, I'm working on it.

4. On my opinion, If the rocks dropped on your toe, it will hurt you less because gravity lessen or like absorbing the impact of the drop. P.S. I guess if it has a possibility to hurt you, why don't you dodge it, you have plenty of time before the rock hit your toe.

Feel free to comment if my Opinion was wrong.

This link might help you to understand about Universal Gravitation: I don't know how to link that one, like when you click it, it will automatically go to the URL. Maybe just highlight, Copy and Paste it. ( You know what to do ^_^ ).

http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

http://www.youtube.com/watch?v=0rocNtnD-yI&feature=related

By the way, April 22 was the due date of LAB WORKSHEET.
Next Scribe is KatieN

Universal Gravitation

Wednesday blog:

Sorry i didn't do the blog last Wednesday because I had a fever. By the way, Wednesday blog is about:

1. Continuation of Lab works
2. Answering the Study Guide Chapter 8

and that's all about Wednesday blog. In addition to, the classroom still the same, nothing had change and pretty much noisy just like the way it is ^_^.

Ms. K was still away and Ms. Voon is our substitute teacher.

ORBIT LAB

ok..so jason will be the next scribe...

all we just did today was THE ORBIT lab..


yeap..that's all we did today...THE ORBIT lab..

if u don't know what to do..the procedures are on the green book chapter 8 page..somewhere between 1-100...

ANALYSIS:
1. force is proportional to the inverse square of the distance... F=1/(d)^2
2. i'm still having a problem w/ this one..
3. still haven't done this...

the lab needs to be hand in tom.

ms.k skipped the class today and will not be around 'til friday??????
ms.voon was the substitute...

anddddd......i guess that's all i need to scribe for today...

scribeye: kepler's 3 laws

hi guys . :)

Today we watched a video entitled "Kepler's 3 laws", where we were given a set of questions about and the answers are:

1. Copernican model

2. Mars

3. 8 minutes of arc / 1/4 apparent width of moon

4. Ellipse

5. Yes

6. focus

7. fireplace

8. poverty; illness

9. to see Tycho Brahe @ Denmark and he wanted to use astronomical data.

10. astronomy

11. he stole'em

12. no

13. planets were moving/spinning while orbiting around the sun (i just made that up but i think thats right :D)

14. 900 pages

15. mars is seen in the same position as if you were on the sun or earth.

16. Yes

17. Greeks

18. Yes

19. eccentricity is how flat the ellipse is.

20. Kepler's 3 laws:

• Planets move around the Sun in ellipses, with the Sun at one focus
• The line connecting the Sun to a planet sweeps equal areas in equal times.
• The square of the orbital period of a planet is proportional to the cube (3rd power) of the mean distance from the Sun (or in other words--of the"semi-major axis" of the ellipse, half the sum of smallest and greatest distance from the Sun)

21. Kepler told people's fortunes (astrology)

-- aaaaaand quizzes from last friday were also returned! if you weren't able to check for corrections here are the answers. :)

1. arrowhead facing down for instantaneous velocity and acceleration has an arrow towards the center of the circle.

2. the tension of the second yoyo string is twice the tension of the first one.

3. a. 13 m/s

b. 39 m/s2

4. 0.098 m.

5. up arrow- fair ; down arrow- gravitational force ; left arrow- centripetal force

6. neither.

7. 63.7 J

8. 14 J

9. 13500 J

10. 36 J

11. a. 296 N/m

b. 0.95 J

c. 0.1036 J

12. 3,594 J

13. i didnt get that sorry. :''<

-- and last! tomorrow we're doing a lab activity, so seeyou guys .:)

--next scribe will be, VAMP. goodluck!

Physcis April 15, 2010

Hello everybody !

Today in our class Ms. K was not around and Ms. Voon was our substitute teacher. We went over the worksheet given to us yesterday "Work and Energy: Springs". We answered the Chapter 8 study guide.

Here are the corrections for the “Work and Energy: springs” worksheet.

1. Fg = mg k= F/x
k = mg/x

2. a. k = F/x
= 10N/.132 m
=75.76 N/m

b. F =kx
=3.0 N/m *.55 m
=1.65 N

c. x = F/k
= 20 N/ 3.0 N/m
= 6.7 m

d. k = F/x
= 19.6 N / .04 m
= 490 N/m

3. F = 95 N/m * .025 m
= 23.75 N

4. Es = 1/2kx^2
x =
=
= 0.53 m

5.a. k =F/x
= 56N/ .18 m
= 311 N/m

b.Es = 1/2kx^2
= 1/2 *311 N/m * (0.18 m)^2
= 5.04 J

6. Es = 1/2kx^2
= 1/2* 144 N/m * (0.165 m)^2
= 1.96 J

7.a. k = F/x
= 12 N/ .6 m
= 20 N/ m



Tomorrow we will have test in Circular motion and Work/ Energy. So guys please stuDY? LOL wahhahaha. Goodluck to all. :)

Camille will be the next scribe.

Physics:)

Hello ladies and gents...

We did answer the Study Guide Chapter 11 which is the "Energy in its Forms" and the "Conservation of Energy." We also went over on the assignment from the green book that Ms. Kozoriz gave us yesterday.

Here are the solutions in the 7 questions in Green Book:

1) KE= 1/2mv^2
= 1/2(1600kg)(12.5m/s^2)
= 1.25x10^5J

2) KE= 1/2mv^2
*108km/h= 30m/s^2

= 1/2(1500)(30m/s^2)
=6.75x10^5J

6 a)v2^2=v1^2+2ad
62m/s^2=0^2+2a(22.0m)
3844m/s=2a(22.0m)
a = 87.36m/s^2

6 b)KE= 1/2mv^2
=1/2(40)(62)^2
=76880J

12 PE= mgh
= 60kg (9.8m/s^2)(3.5m)
= 2058J or 2.1x10^3J

13) PE= mgh
= 64kg (9.8m/s^2)(2.1m)
= 131.72J

20 a) W=F.d
=98N (50m)
=4900N.m

20 b) PE= mgh
= Fh
=98N (50m)
=4900J

20 c) W= KE
= F.d
=98N(50m)
=4900N.m

22 a) KE= F.d
= 1/2mv^2
= 201kg(1.3m/s)
1/2 = mv^2 =261.3
v2 = 261.3/.15
= 1742m/s(Find the square root)
= 41.73 or 42m/s

20b) v2^2=v1^2+2ad
d =0-(42^2)/2(9.8m/s)
=90m

After the discussion, Ms. Kozoriz gave us another work sheet which is the "Work and Energy: Springs"



Jape is the next scriber...

Physics!!!

Hi!

Today in class we went over and answered the three sheets, Chapter 10 Study Guide [stapled version], Concept-Development Practice Page [Chapter 8: Energy 8-1 & 8-2].

Ms.Kozoriz gave us some assignments in green book and a sheet [Chapter 11 Study Guide] that i think we'll correct tomorrow in class.

Next scribe is going to be Rodel.

Physics!

Hi everyone :)

Today in class we went over the booklet, Work and Energy. The notes basically introduce different concepts of "work." We also corrected Chapter 10 Study Guide: 10.1 Work and Energy. Here are the answers:
Work
force, distance, direction
W=FxD
work, force, distance
scalar
joule
Newton, meter, joule
moves
force, displacement
Work and Direction of Force
in the direction of
perpendicular
in the direction of
cosine, direction of motion
negative
energy
energy, motion
positive, negative
Power
rate, rate, energy
P=W/T
power, work, time
watt
joule, one second
kilowatts

Corrections were also done for p.330 of the Physics book, answers are as follows..
1.a) W=FD
=(40N)(0.15m)
=6J

b) W=FD
=(50kg)(9.8)(195m)
=956J

c) W=F∆D(cosΘ)
=(120)(4.0)(cos25)
=435J

2) W=FD
=(5000)[1/2(25)5^2]
=156 250J

3) W=F∆D(cosΘ)
=(78N)(10m)(cos35)
=447J

I wrote down the answers for questions 4-5 but, I wasn't sure of the answers! Sorry =/

6) D=W/F
=480J/30N
=16m

Ms.K also handed back our Centripetal Force Labs back, here are the answers to the questions in case you didn't get them. Thanks to group mates Dave and Camy for getting the correct answers..nerds❤ ..or i'm just dumb =S
2) The relationship revealed is a power relationship between the two variables.
4) A straight line is obtained, representing a linear relationship between F&V^2.
5) F=mv^2/R... m(mass) and R(radius) were kept constant throughout the investigation. Therefore yes, the results of our investigation do support the equation because F over V^2, representing a power relationship as displayed by the results.

Homework:

• Chapter 11 Study Guide: 11.1 Energy in its Many Forms
• 8-1, 8-2 Concept-Development Practice Page

Zaiza to scribe next.

Another day in physics

Hello phsyics bloggers im gonna make this sweet and short cause im cool like that.

Anyways! Today in class we went over questions 13 to 18 on pages 152 & 153 in the green text books. I would have posted the answers for people WHO did not have the time or forgot to copy the answers but lynel forgot to give me the papers. we also started another lab (Your Power) near the end of the period but thats more for tomorow..

So thats all for now, ill be back when someone makes me scribe again haha i do a horrible job so i doubt that will happen. okay byee!

The next scribe is Lynel Pobre

CENTRIPETAL FORCE LAB

Hello everybody!

Today was pretty simple. All we did was work on our Centripetal Force lab and if you missed today's class make sure you talk to Mrs.K!

Next scribe is going to be Roger.

First Day Back From Spring

Hey. How was everyone's Spring Break. Did everyone have fun because I know what I did.
Okay, so today in class we went over the Momentum and Impulse test. The answers are posted underneath my post thanks to Ms. K. And after it was either you do the lab since some people are not going to be here on Wednesday or do textbook questions.
-For the lab, it is due on Thursday as what Ms. K told me and the lab is on circular motion. If you get the lab sheet, it explains on what to do. It is fairly easy.
-For the textbook questions, it is on Pg. 152 questions number 15-18. They are practice questions on circular motion.
And that is all for today's class.
The next scribe is going to be Daniel Omana.

Momentum and Projectile Motion Test Answers

Momprojtest

View more presentations from ekozoriz.

some infors about circular motion

Circular Motion
A body that travels an equal distances in equal amounts of time along a circular path has a constant speed but not constant velocity. This is because velocity is a vector and thus it has magnitude as well as direction
The velocity of P is directed along the tangent at P. The speed remains constant but the velocity has changed. We know that if the velocity changes with time then the ball on the string is also accelerating.
The Radian
angles can be measured in radians as well as degrees. The angle in radians is defined by. If s = r then θ=1 rad. Therefore, 1 rad is the angle subtend at the center of a circle by an arc equal in length to the radius. When s =2π r then θ=2 π radians=360°. Therefore, 1 rad = 360°/2 π=57.3°
θ = s/r(1)
Since we are often dealing with angles and trigonomic functions, a useful approximation that is often used is that for small angles of &theta, sin θ and tan θ = θ where θ is measured in radians.
&theta = sin &theta = tan θ for small angles where theta is measured in radians.
Angular Velocity
The speed of a body moving in a circle can be specified either by its speed along the tangent at any instant (linear speed) or by the angular velocity. This is the angle swept out in unit time by the radius joining the body to the centre. It is measured in [rad s-1]
Further, the angular velocity is the time for the particle to travel around once divided by the period. ω = 2 π/T
ω = θ/t(2)
Derivation
Consider a body moving uniformly from A to B in time t so that OA rotates through a small angle θ.The angular velocity, ω of the body about O is ω=θ/t.If arc AB has length s and v is the constant speed of the body, v=s/tBut s=r θ.So, v=r θ/t and ω =θ/t,therefore v=rω
v = r/ω(3)
Acceleration
The expression for the acceleration of an object moving in circular motion of radius r moving at a constant speedv is derived as follows. If it travels from A to B in a short interval of time δt then, since speed = distance x time, arc AB = vδt.Also by the definition of angle in radians, arc AB = r δθ=v δt.So, v δt/r = δθ.
The vectors vA and vB represent the velocities at these points. The change in velocity between A and B obtained by subracting vA from vB.
Change in velocity = vA and vB. Parallagram rule or triangle rule. The resultant is effectively XZ. Since one vector (-vA) is perpendicular to OA and the other vB is perpendicular to OA and the other vB is perpendicular to OB, therefore ∠ XYZ = ∠ AOB =δθ
If δt is small, the δθ is also small. Then the length of XZ in the figure will be almost the same as X'Y' in b) X'Z'=rδθ but also δθ=vδt/r. Therefore, X'Z'=v2/r δt.
a=change in velocity/time interval = (v2δt/r)/δt=v2/r. Therefore a=v2/r also a=ω2r.
a = v2/r(4)
a = r ω2(5)
The direction of the acceleration is toward the centre O as can be seen if δt is made so that A and B all but coincide; XZ is then perpendicular to vA (or vB) is along AO or BO. We say the body has a centripetal acceleration (i.e. centre seeking). Does a body moving uniformly in a circle have constant acceleration?
Centripetal Force
Since a body moving in a circle (or circular arc) is accelerating, it follows that from Newton's 2nd law that there must be a force acting on it to cause the acceleration. This force will also be directed toward the centre and is called the centripetal force. It causes the body to deviate from the straight line motion it would naturally follow. The magnitude of the centripetal force is given by:
F = ma = mv2/r(6)
Where m is the mass of the body and v is the speed in the circular path of radius r. If the angular velocity of the body is ω we can also say since v=rω.
F = ma = mrω2(7)
Other examples of circular motion will be discussed in the following sections. In all cases it is important to appreciate that the forces acting on the body must provide a resultant force of magnitude mv2/r toward the centre.
Experimental Measurement of Centripetal Force
The turntable is rotated by the electric motor causing the truck of known mass m to move out from the centre of the turntable. Opposing the motion of the truck is a spring which provides the centripetal force required to keep the truck in position. As the speed of the turntable is increased the spring extends until the truck reaches the stop at the end of the rails. When the truck just hits the truck the extension of the spring is of known length. The velocity of the turntable can be found by measuring the time it takes for the turntable make ten revolutions and dividing by ten. The velocity it then 2π/t. We know the m and the radius and therefore the extension of the spring. We can also measure the force required to produce the same extension of the same using a spring balance and compare the two forces.
Rounding A Bend
If a car is travelling round a circular bend with uniform speed on a horizontal road, the resultant force acting on it must be directed to the centre. ie. it must be the centripetal force. This force arise from the interaction of the car with the air and the road. The direction of the force exerted by the air will more or less oppose the instantaneous direction of motion. The other more important horizontal force is the frictional force exerted inwards by the road on the tyres of the car. The resultant of these two forces is the centripetal force. If the centripetal force is less than the force wanting to pull the car out, a skid will result. Safe cornering that does not rely on friction is achieved by banking the road. The problem is to find the angle θ at which the bend should be banked so the centripetal force acting on the car arises entirely from a component of the normal force, N of the road.
Treating the car as a particle and resolving N vertically and horizontally we have, since N sinθ is the centripetal force. N sinθ=mv2/r where m and v are the mass, speed and r is the radius of the bend respectively. Also the car is assumed to remain in the same horizontal plane therefore no horizontal acceleration therefore, N cosθ=mg. Dividing both equations, tan θ=v2/rg
A bend in a railway track is also banked. In this case, so that at a certain speed no lateral thrust has to be exerted by the outer rail on the flanges of the wheels of the train, otherwise the rails are strained. The horizontal component of the normal forces of the rails on the train provide the centripetal force.
Aircraft

Figure. A banking aircraft uses the horizontal component of the lift force to provide the centripetal force for turning.
An aircraft in straight, level flight experiences a lift force perpendicular to the surface of its wings which balances its weight, mg. To make a turn, the ailerons are operated so that the aircraft banks and the horizontal component of the lift supplies the necessary centripetal force to make the aircraft turn. The lift force is always perpendicular to the wings and so the aircraft's weight has to be supported by the vertical component of the lift. Since this is slightly less than the weight of the aircraft, the aircraft will lose altitude unless the lift is increased by increasing the speed or pulling the nose of the aircraft up.
The Rotor
The rotor is often seen at a fairground. A cylinder of around 4 meters in diameter is set spinning. The passengers stand inside the cylinder against the wall as it spins. The rotational motion causes the passengers to be pinned to the wall of the rotor. When the rotor is spinning fast enough, the floor of the cylinder drops down and the passengers remain pinned to the walls. The forces acting on a passenger of m are shown. N is the normal force and is the centripetal force needed to keep him moving in a circle. Hence if r is the radius of the rotor and v the speed of the passengers then N=mv2/r. If F is the frictional force acting upwards between the passenger and the rotor wall and since there is no vertical motion of the passenger. F=mg. If μ is the coefficient of limiting friction between the passenger and the wall we have F=μN. Therefore, μN=mg, μ=mg/N=gr/v2. Note that this is independent of the mass m.
A typical value of μ between clothing and the rotor wall (canvas lined) is about 0.40 so if r=2m. v=5ms-1. What would be the angular velocity?
Since v=rω, &omega=7/2 =3.5 rad s-1. Revolutions per minute = (3.5 rads-1x60 sec)/2π rad=210/2π=105/π
Looping the Loop
Consider a bucket of water when it is at the top of the loop A. If the weight of the water mg is less than mv2/r, the normal force N of the bottom of the bucket on the water. However, if the bucket is swung more slowly than mg will mv2/r and the residual part of the weight causes the water to leave the bucket.
Centrifuge and Washing Machines
Clothes get damaged in the washingmachine because the clothes experience a high force of gravity
Space Elevator
One application of centripetal force and circular motion is found in the space elevator lift. It was proposed by Russian scientist, Yuri Artsutanov in 1960 as a possible way of getting into space. It works on the principle that at a object orbiting the Earth above the equator with a period of 24 hours will remain in the same position above the Earth (see geostationary satelites). If a rope were to be lowered from the object, then an elevator car could climb up the rope into space without the need for a rocket. Since the rope has mass the additional weight of the rope would cause the satellite holding it to be pulled back to Earth. However, a counter weight orbiting the Earth a greater distance than the geostationary height would be travelling faster to stay in the same position over the equator the centripetal force of the counterweight can be balanced against the weight of the rope. The result is that the rope stays in place and the space elevator becomes possible.
Even though the theory is sound, there are significant technical challenges yet to be overcome before the space lift becomes a reality. The first is to do with the material properties of the rope. Given that the height of the geostationary orbit is some 3.58 x 104 km above the equator, any rope would have to be extremely strong to support its own weight. At present there is no known material that would be able to support its own mass at this length however, carbon-nanotubes have been suggested as a possible material for such ropes because of their low mass per unit length and extreme strength.
Other problems such as space debris also need to be overcome.


all the infos above from a useful website
http://www.splung.com/content/sid/2/page/circular
so if you guy want to know more details about this chapter, you can click the link above.