Hello everyone! I'm Tim, your scribe for Friday. As much as I would like to make this a quality scribe, there isn't much that I can say that we've completed on Friday, considering the fact that a huge portion of our class (4th period) was consumed by BUS RIDERSHIP! Thank goodness we only have one additional bus ridership left this year.
During the Period, two main things happened. The first was that we went over questions that involved friction, inclined planes, gravitational force, normal force, parallel + perpendicular forces, and all of that fun stuff. These questions were on the worksheets handed out to us.
Next, we worked on a handout that involved looking at vectors. Observing their characteristics, we found the vector resolutions by utilizing the vector components. Remember to break the vectors into perpendicular components before solving (x and y components). To solve for individual components, you have a number of strategies at your disposal.. including the Pythagorean Theoream and Trigonometry.
Well that's the brief runthrough of Fridays class! See everyone at school tomorrow!
Monday's Scribe is... Sharmaine D??
Off to H3!
Sunday, September 30, 2007
Thursday, September 27, 2007
Review Time
Hello fellow students of the Physics which Ms. K teaches. Today in class we went over some of the questions on the hand outs that we were given and gotten the answers for most of them. Basically today was just a review because tomorrow we don't have a physics class, well, for the 3rd periods , we do not have a class all because of pep-rally ! WHOOOT ! That was the majority of our class soo, STUDY HARD KIDS ! TEST is on MONDAY ! however some people do forget what was going on during class within 3 days away from doing physics, so I suggest you all look over your notes a couple, or maybe a bajillion times to ensure that your brain has functioned correctly and have stored the information it needs for the physics test on Monday. Take note that there will not be any of the Pulley questions on the test.
Next Scribe is Tim_MATH_y !!!
Next Scribe is Tim_MATH_y !!!
Wednesday, September 26, 2007
In class work time
hey its Anthony. I will be the scribe for today. I don't really know what to say. Today's class pretty much just slipped by without me learning anything new. The sub Mr. Whiteway just sat upfront and watched over us. We got two new worksheets to practice on. "advanced forces" and "Enrichment chapter 5". Alright test is on monday, remember to study. Also there is a question worth 5 bonus marks on mondays test to bring up your previous mark if you didn't do so well.
next scribe is OLIVER
next scribe is OLIVER
Tuesday, September 25, 2007
Initiation of Review
Hi, I'm Chris (from third period physics) and I will be the scribe for today. I simply titled the post "initiation of review" since all we really covered today was a small review of some sample problems involving forces and free body diagrams. Throughout the class, Mrs. K attempted to detail each of the presented cases (found within a small booklet that was also given out today) sufficiently so that the entire class could comprehend the problems at hand. Though, I'm not so sure everyone in the class was quite as comfortable with these cases as she might have thought. In this post, I will outline each of the cases that were covered in today's class with hopefully proficient detail and explanation that the class might delve into the monstrous test on monday with some apt confidence and security.
________________________________________________________
Case 1: Constant Acceleration Motion
a) Ff
b) Fnet
c) a"
Now, this is probably one of the simplest dynamics question that we might encounter on the test on monday.
For part (a), the question asks for us to find Ff. Simple enough, we have been taught that to calculate Ff, the equation would be as follows:
Ff = μFn ; where Ff = friction, μ=coefficient of friction and Fn = normal force.
Now since we know that Fn = -Fg, since the normal force is equal and opposite to the weight of the object, we can then substitute Fg into the equation instead of Fn. We would do so because we know the mass of the object, we can then substitute m*g instead of Fg to determine the weight or normal force (since they are equal in linear dynamics), then we would just multiply it by the coefficient of friction to determine the frictional force. This can be shown algebraically as follows:
Ff = μFg = μmg = (0.15)(50 kg)(-9.81 N/kg) = -73.5 N (negative since it is moving left)
For part (b), we are looking for the net force in the free body diagram, and since we know that Fn and Fg are equal, we do not need to take them into consideration. We can also determine from the free body diagram, and also since there is acceleration occurring, is that the applied force is greater than the frictional force. So in order to determine the net force, we must add up the forces Fa and Ff. Since we know both, this portion becomes quite easy.
Fnet = Fa + Ff = (100 N) + (-73.5N) = 26.5 N (since it's positive, it's moving right)
For part (c), the accceleration can be determined from rearranging the equation Fnet = ma to solve for a:
a = Fnet / m = 26.5 N / 50kg = 0.53 m/s2
_________________________________________________________
Case Two: Force on Two Masses
"If the applied force is 200 N, m1 = 5 kg and m2 = 3 kg, and μ = 0.1, calculate
a) Fnet
b) a
c) F2"
Now this question presented tribulations for the class as Mrs. K attempted to present how to approach this problem to the class.
For part (a), to calculate the net force on the objects, all we must do is add Ff and Fa together (since once again, Fn = Fg). Now, we already know Fa, but we then need to determine Ff. Since Ff = μmg, as I outlined earlier, we need to apply that once again but instead of having just one mass, we have two, so we must combine the two masses to get m. By this I mean that in this instant we treat the two masses as one mass by combining them into one. This will give us:
Fnet = Ff + Fa = μ(m1+ m2)g + Fa = (0.1)(5 kg + 3 kg)(-9.81 N/kg) + 200N = 192 N
Now for part (b), we are attempting to determine the acceleration of the two masses. All we must do for this is use the net force (since it is the total of all forces in the situation) and divide it by the total mass of the object, once again adding m1 + m2. Algebraically:
a = Fnet / (m1+m2) = 192 N / 8 kg = 24 m/s2
Part (c) requests that we now determine F2, this usually presented trouble for some students so I will try my best to outline the process of calculating F2. In order to calculate this force value, we must see that F2 is caused by mass 2 and is actually driving m2 forward, but is pushing back on m1. This is a direct result of Newton's third law, that every action will have an equal and opposite reaction. We must also realize that this force is a ratio of the overall force being applied, in other words, it is a specific portion of the 200 N being applied to the overall object (m1 and m2 together). To determine this so-called ratio, we must divide the mass of m2 by the total mass, since we're basically looking for how much of the total object is just m2. Now, here's how the calculations should look like:
F2 = [m2 / (m1 + m2)] F = (3/8) (200 N) = 75 N
________________________________________________________
Case Three: Connected Masses
a) a
b) T1
c) T2"
To determine the acceleration, since there is no friction acting (and Fg = Fn, so they quantitatively cancel each other) Fnet = F. Also, in this case, we must incorporate all three masses into our calculations using m since F is indirectly acting (pulling) the entire connection of the masses together. Now, to calculate acceleration, we can use the formula F = ma rearranged to solve for acceleration:
a = F / (m1 + m2 + m3) = 10N / 7 kg = 1.43 m/s2
To determine the first tension force (acting on m1 and m2), we can use the first free body diagram to carry out this calculation, seeing as it is more simple. Now, as we inspect the first diagram, we notice that the only force acting on the object IS T1, meaning that the net force in this case IS T1. So, in order to calculate T1, we can use the equation T1 = m1a.
T1 = m1a = (2 kg)(1.43 N/kg) = 2.86 N
For T2, we must use the third free body diagram to calculate it since we can use F as a constituent of the net force to determine T2. This, in other words, means that we can use the net force (the total of the forces) to determine T2 since F is known. The vector T2 would be negative in this situation, since it is moving left, and F would be positive as it would be moving right. This doesn't always have to be the case, but I have designated left negative and right positive, or the force T2 negative with respect to F. Since the net force is not known, but m3 and a are both known, we can use those values in place of Fnet since Fnet = m3a. This gives us:
T2 --> F - T2 = m3a --> T2 =-(F - m3a) --> T2 = -[10N - (1 kg)(1.43 N/kg)] = -8.57 N
____________________________________________________
Cases Four and Five:Mass on an Incline
- If Fa is less than Ff, then the object is not moving.
- If Fa = Ff, then the object is moving with a constant velocity.
- If Fa > Ff, then the object is accelerating.
Continuing with the explanation of frictional force in an inclined plane is imminently complete. As I noted earlier, Ff opposes F parallel, but also, in the case of all situations regarding frictional forces there underlies a specific coefficient of friction (the ratio of friction over the normal force). Using either of these two facts leads to the calculation of the frictional force acting on an object on an inclined plane. To use these two ways, either subtract the F parallel force from the net force to get the frictional force, or manipulate/directly use (depending on for what purpose) the equation Ff = μFn.
________________________________________________________
Now, I hope many of you, especially those who were absent from today's class, have acquired more of a formally but yet simplified (to an extent) explanation of some of the main concepts and processes within this unit.
As our class finished covering the above through a somewhat less elaborate volubility of explanation, we were given two sheets entitled "C. Using Concepts" and "Grade 12 Physics: Dynamics Problems" to complete. Though I am not sure whether they are for homework in the sense that we will correct them or hand them in or not, I strongly encourage many to not only attempt and complete the questions, but to shamelessly inquire as necessary. These questions will not only provide more practice and a more definitive grasping of the subject at hand, but also a stronger habitual behaviour to approach problems that we might encounter on the upcoming test. As for the test, it will be on Monday, though some argue (can't remember who else besides Vincent) that the test should be on Tuesday. That is all for my scribe post, I know it was lengthy, albeit an obvious resultant of my instructional endeavor for the rest of the class(es). Hope this post can help anyone at all in this unit, have a great night and don't forget to do homework!
Oh yes, the scribe for tomorrow will be Anthony!
________________________________________________________
Case 1: Constant Acceleration Motion
a) Ff
b) Fnet
c) a"
Now, this is probably one of the simplest dynamics question that we might encounter on the test on monday.
For part (a), the question asks for us to find Ff. Simple enough, we have been taught that to calculate Ff, the equation would be as follows:
Ff = μFn ; where Ff = friction, μ=coefficient of friction and Fn = normal force.
Now since we know that Fn = -Fg, since the normal force is equal and opposite to the weight of the object, we can then substitute Fg into the equation instead of Fn. We would do so because we know the mass of the object, we can then substitute m*g instead of Fg to determine the weight or normal force (since they are equal in linear dynamics), then we would just multiply it by the coefficient of friction to determine the frictional force. This can be shown algebraically as follows:
Ff = μFg = μmg = (0.15)(50 kg)(-9.81 N/kg) = -73.5 N (negative since it is moving left)
For part (b), we are looking for the net force in the free body diagram, and since we know that Fn and Fg are equal, we do not need to take them into consideration. We can also determine from the free body diagram, and also since there is acceleration occurring, is that the applied force is greater than the frictional force. So in order to determine the net force, we must add up the forces Fa and Ff. Since we know both, this portion becomes quite easy.
Fnet = Fa + Ff = (100 N) + (-73.5N) = 26.5 N (since it's positive, it's moving right)
For part (c), the accceleration can be determined from rearranging the equation Fnet = ma to solve for a:
a = Fnet / m = 26.5 N / 50kg = 0.53 m/s2
_________________________________________________________
Case Two: Force on Two Masses
"If the applied force is 200 N, m1 = 5 kg and m2 = 3 kg, and μ = 0.1, calculate
a) Fnet
b) a
c) F2"
Now this question presented tribulations for the class as Mrs. K attempted to present how to approach this problem to the class.
For part (a), to calculate the net force on the objects, all we must do is add Ff and Fa together (since once again, Fn = Fg). Now, we already know Fa, but we then need to determine Ff. Since Ff = μmg, as I outlined earlier, we need to apply that once again but instead of having just one mass, we have two, so we must combine the two masses to get m. By this I mean that in this instant we treat the two masses as one mass by combining them into one. This will give us:
Fnet = Ff + Fa = μ(m1+ m2)g + Fa = (0.1)(5 kg + 3 kg)(-9.81 N/kg) + 200N = 192 N
Now for part (b), we are attempting to determine the acceleration of the two masses. All we must do for this is use the net force (since it is the total of all forces in the situation) and divide it by the total mass of the object, once again adding m1 + m2. Algebraically:
a = Fnet / (m1+m2) = 192 N / 8 kg = 24 m/s2
Part (c) requests that we now determine F2, this usually presented trouble for some students so I will try my best to outline the process of calculating F2. In order to calculate this force value, we must see that F2 is caused by mass 2 and is actually driving m2 forward, but is pushing back on m1. This is a direct result of Newton's third law, that every action will have an equal and opposite reaction. We must also realize that this force is a ratio of the overall force being applied, in other words, it is a specific portion of the 200 N being applied to the overall object (m1 and m2 together). To determine this so-called ratio, we must divide the mass of m2 by the total mass, since we're basically looking for how much of the total object is just m2. Now, here's how the calculations should look like:
F2 = [m2 / (m1 + m2)] F = (3/8) (200 N) = 75 N
________________________________________________________
Case Three: Connected Masses
a) a
b) T1
c) T2"
To determine the acceleration, since there is no friction acting (and Fg = Fn, so they quantitatively cancel each other) Fnet = F. Also, in this case, we must incorporate all three masses into our calculations using m since F is indirectly acting (pulling) the entire connection of the masses together. Now, to calculate acceleration, we can use the formula F = ma rearranged to solve for acceleration:
a = F / (m1 + m2 + m3) = 10N / 7 kg = 1.43 m/s2
To determine the first tension force (acting on m1 and m2), we can use the first free body diagram to carry out this calculation, seeing as it is more simple. Now, as we inspect the first diagram, we notice that the only force acting on the object IS T1, meaning that the net force in this case IS T1. So, in order to calculate T1, we can use the equation T1 = m1a.
T1 = m1a = (2 kg)(1.43 N/kg) = 2.86 N
For T2, we must use the third free body diagram to calculate it since we can use F as a constituent of the net force to determine T2. This, in other words, means that we can use the net force (the total of the forces) to determine T2 since F is known. The vector T2 would be negative in this situation, since it is moving left, and F would be positive as it would be moving right. This doesn't always have to be the case, but I have designated left negative and right positive, or the force T2 negative with respect to F. Since the net force is not known, but m3 and a are both known, we can use those values in place of Fnet since Fnet = m3a. This gives us:
T2 --> F - T2 = m3a --> T2 =-(F - m3a) --> T2 = -[10N - (1 kg)(1.43 N/kg)] = -8.57 N
____________________________________________________
Cases Four and Five:Mass on an Incline
Now, for these cases (though we did not go over them in class, but were told to complete them) I will outline how to come about solving such questions. These inhabited a majority of the current unit and is quite likely to be found frequently throughout the upcoming test. Now, the following diagram will be used to describe both scenarios.
The normal force, Fn, will always act perpendicular to the surface. This meaning that if the mass is resting on a surface, the normal force exerted on the mass will form a 90o with the surface. Cases which involve an object and an incline will always direct gravity downwards, regardless of the slope of the surface, the direction of the normal force, etc. Gravity is always pointed down (as shown in the diagram) in an inclined plane problem. Now, in order to conduct calculations, we must break the force of gravity, Fg, or weight (all the same thing), of the object into two components, a parallel component and a perpendicular component. These are basically just x and y components (or vertical and horizontal components) of the force of gravity, but since Fg is now acting as the hypotenuse pointed directly down, they are named according to their position in the diagram. By this I mean that these components are named relative to the surface, as with in the case of the normal force. These components are illustrated in the freebody diagram depcited above. To calculate these components, we visualize the triangle formed by breaking Fg into it's components as a right triangle, where Fg itself is the hypotenuse. The angle θ will be formed between Fg and F perpendicular, and from there on use basic trigonometric functions (particularly sinθ and cosθ) to determine either F perpendicular or F parallel. And if Fg is not given, but the mass is, then mg can always be used to substitute for Fg.
- If Fa is less than Ff, then the object is not moving.
- If Fa = Ff, then the object is moving with a constant velocity.
- If Fa > Ff, then the object is accelerating.
Continuing with the explanation of frictional force in an inclined plane is imminently complete. As I noted earlier, Ff opposes F parallel, but also, in the case of all situations regarding frictional forces there underlies a specific coefficient of friction (the ratio of friction over the normal force). Using either of these two facts leads to the calculation of the frictional force acting on an object on an inclined plane. To use these two ways, either subtract the F parallel force from the net force to get the frictional force, or manipulate/directly use (depending on for what purpose) the equation Ff = μFn.
________________________________________________________
Now, I hope many of you, especially those who were absent from today's class, have acquired more of a formally but yet simplified (to an extent) explanation of some of the main concepts and processes within this unit.
As our class finished covering the above through a somewhat less elaborate volubility of explanation, we were given two sheets entitled "C. Using Concepts" and "Grade 12 Physics: Dynamics Problems" to complete. Though I am not sure whether they are for homework in the sense that we will correct them or hand them in or not, I strongly encourage many to not only attempt and complete the questions, but to shamelessly inquire as necessary. These questions will not only provide more practice and a more definitive grasping of the subject at hand, but also a stronger habitual behaviour to approach problems that we might encounter on the upcoming test. As for the test, it will be on Monday, though some argue (can't remember who else besides Vincent) that the test should be on Tuesday. That is all for my scribe post, I know it was lengthy, albeit an obvious resultant of my instructional endeavor for the rest of the class(es). Hope this post can help anyone at all in this unit, have a great night and don't forget to do homework!
Oh yes, the scribe for tomorrow will be Anthony!
Monday, September 24, 2007
Friction Lab Analysis (Sept 24)
Today we finished up the Friction Lab which involved a block of wood and its measured movement on an inclined plain. After retrieving the data from the experiment we were asked to use that data to fill out a table on the forces which were acting on the block as it climbed the plain at constant velocity. We were also given the answers to Chapter 5 Review (shown below).
1. 0.3m/s²
2. 5.0kg
3. 2.2 x 10^4N
4. 144N
5. -0.53m/s²
6. 670N
7. 9.0m/s²
8. 0.46
9. 188N
10. 7.1N
11. 384N
12. F= -820N
Chris is the new scribe....muahahahahahahahaha.....quiet you!
1. 0.3m/s²
2. 5.0kg
3. 2.2 x 10^4N
4. 144N
5. -0.53m/s²
6. 670N
7. 9.0m/s²
8. 0.46
9. 188N
10. 7.1N
11. 384N
12. F= -820N
Chris is the new scribe....muahahahahahahahaha.....quiet you!
Friday, September 21, 2007
Scribe Post
Well first time this year I am scribing for physics. Today in class we had a lab to complete. The lab was based on dynamic forces which we have been learning for the past week. The lab consisted of a block of wood which was tethered to piece of string on a fly wheel. The block of wood was placed at various degree inclines in which it would be slide down to see what forces were acting on the piece of wood and which force became large enough to move the piece of wood down the incline. The lab is due Monday, as we will be given time in class to finish the lab, and all its questions. Due along with the lab, are the 12 questions Ms. K assigned us on Thursday to be completed for Monday. The next scribe will be ..... Steven F.
Have a good weekend!
Have a good weekend!
Thursday, September 20, 2007
Sept. 20 -- More friction
Ok, so now that I finally got home from practice let's see if I can do this thing rght. So today in class we first went over the last two pages of the study guide and corrected it. We learned that Ms K is weak and cannot overcome the force of friction on a table, and has a hard time overcming the friction on an eraser (lol). Then, thanks to Steven, we got a Dynamics unit sheet that talks about normal force, friction and the inclined plane, which explains when to calculate for friction. Lastly, we got two more sheets to work on, one of which all we could work as a group to answer which had to be handed in for marks, in which all we had to do was answer the questions based on common sense (no calculations), and also the Chapter 5 review. Oh yeah and if anyone has the answers for this last one, please share haha.
So for tomorrow Dino can have the honors.
So for tomorrow Dino can have the honors.
Wednesday, September 19, 2007
Scribe...
Hello everyone,
Today we started off by doing questions that we were assigned a while back. Kasia and Craig took the reigns there...
After that Mrs.K showed us how vectors work on angled planes. Basically you just have to separate gravity into parallel and perpendicular forces and your good...
Then we were assigned a quick review read from our texts and then a quicker section on angled planes...
We have to finish a sheet that was handed out and that other assignment for tomorrow...
The next scribe will be RENAN!
Have a good night,
cya
Today we started off by doing questions that we were assigned a while back. Kasia and Craig took the reigns there...
After that Mrs.K showed us how vectors work on angled planes. Basically you just have to separate gravity into parallel and perpendicular forces and your good...
Then we were assigned a quick review read from our texts and then a quicker section on angled planes...
We have to finish a sheet that was handed out and that other assignment for tomorrow...
The next scribe will be RENAN!
Have a good night,
cya
Tuesday, September 18, 2007
Dynamics//Equilibrium
Hello hello everyone, this is Kim and I'll be doing our little scribe post for today.
Today we got a handout that went over our dynamics unit. It defined equilibrium (an object is in equilibrium when it has zero acceleration) and it also took us through a reasoning strategy for analyzing equilibrium situations. The steps are as follows:
Today we got a handout that went over our dynamics unit. It defined equilibrium (an object is in equilibrium when it has zero acceleration) and it also took us through a reasoning strategy for analyzing equilibrium situations. The steps are as follows:
- Select the object to be studied.
- Draw a "free-body diagram."
- Choose a set of x and y axes.
- Set up equations so that the sums of the x and y axes are zero.
- Solve the equations for the unknown quantities.
These five steps should be utilized when solving equilibrium problems in 2D (the more complicated ones).
In the case below (1D) the problem is made much simpler because all you need to solve the problem is the mass of the object. Using the mass, you can plug it into the formula, F=mg. g=-9.81 and multiplying this force of gravity to the mass you will find the downwards force (F).
Once you discover the value, it's simply a matter of realizing that the downwards force you already solved for is equal and opposite to the upwards force.
Thus, the values for each of the 2 cables holding the sign (T) are simply half of the upwards force.
**Note: Ms. K gave us all an assignment on static equilibrium that will be due tomorrow!
Personally, with plenty of help and explanations, I'm slowly understanding all of the concepts we're dealing with and the best thing to do if you don't understand is to simply ask questions. Ask Ms. K, your classmates or anyone who can help you.
Have a good night everyone and don't forget your assignments since it is for marks.
Monday, September 17, 2007
Tests given back!
Well it's Nicole because Kim had to work today so she will be doing the post for tomorrow, so I will begin.
Today we got our tests back once we all sat down and for most of us it wasn't exactly the greatest. We then went over the answers that I am sure Ms. Kozoriz will be posting later.
We then got some important defenitions, which were:
Dynamics - motion and forces
Statics - all forces are balanced to keep objects still
Static equilibrium - object's state of no motion when all the forces acting on it are balanced
Dynamic equilibrium - not moving or at constant acceleration
We also went over questions 1-3 on p.237 from Friday's class. The answers were:
1. adj= 5000N opp=8660
2. 68.4N
3. a) free body diagram
b) 0.039m
c) 0.45kg
We were then told to do the first page of Ch. 7 Study Guide and finish up questions #1-3 on p. 266 and #18-20 on p.267. So that was every for today and tomorrow it will be KIM!
Today we got our tests back once we all sat down and for most of us it wasn't exactly the greatest. We then went over the answers that I am sure Ms. Kozoriz will be posting later.
We then got some important defenitions, which were:
Dynamics - motion and forces
Statics - all forces are balanced to keep objects still
Static equilibrium - object's state of no motion when all the forces acting on it are balanced
Dynamic equilibrium - not moving or at constant acceleration
We also went over questions 1-3 on p.237 from Friday's class. The answers were:
1. adj= 5000N opp=8660
2. 68.4N
3. a) free body diagram
b) 0.039m
c) 0.45kg
We were then told to do the first page of Ch. 7 Study Guide and finish up questions #1-3 on p. 266 and #18-20 on p.267. So that was every for today and tomorrow it will be KIM!
Sunday, September 16, 2007
The National Scribe Post.
Hey there, it's Kasia's turn! So on friday we walked into our physics class to a surprise unlike any other, Mrs. Kozoriz was missing! Fortunately we had a sub who gave us a multiple of names to call him by, such as, Mr. Reimer, Kelly, Kelly R., R-Kelly. Then he proceeded to give us our assignment for the day. We were to read pages 232 7.1 and 234-37 7.3 from the duck book. That, of course, didn't take all period so we were to do questions 1-3 on page 237, 1-3 on page 266 and 18-20 on page 267, that did take up the whole class. One question in general seemed to be giving the majority of the class the most trouble and that was question 2 on page 237. So that is a heads up to anyone coughMrs.K.cough who missed that day and might feel that the question might need going over on Monday. Basically that was the class in a nutshell guys.
And now the moment you've all been waiting for.....the next scribe is............................................................KIM!
And now the moment you've all been waiting for.....the next scribe is............................................................KIM!
Thursday, September 13, 2007
Test Scribe...
Well, I said I wouldn't be able to post until very late tonight, but I found time during my spare, sooooooo I figure I might as well do it now. Today's class consisted of a test, and... that's about it. It was fairly easy, excpet the last couple of questions were a little more difficult. The 5th question, BTW has two possible ways of drawing the diagram, one makes more sense than the other, but that's besides the point. This means that there are two answers that could be deemed correct, but it dpends on how Mrs. K marks it. I think the one that I got was 19 km/h @ 22 W of N and Mr.s K. said she used that method as well. The other, was what Chris got, but i can't quite remember what it is. Anyway, I guess we'll have to see how that plays out.
Next Scribe is KASIA.
See you all tomorrow =D
Next Scribe is KASIA.
See you all tomorrow =D
Wednesday, September 12, 2007
RELATIVE MOTION in Two Dimensions
TODAY, we went over the questions 37 - 42 from the textbook. There were some discussions about each question and the method used to come up with the right answer. Designing the diagram was a bit of a problem for some and that sort of messed up some people's answers. See slides posted by Mrs. Kozoriz to view the solutions and answers.
After that, we were given another sheet to work on. I'm sure you all have it so I'm just going to post the right answers I took down.
VECTOR COMPONENTS:
1.a) 8660 km
b) 5000 km
2) 192 Newtons
3) 326 Newtons
4) 192 seconds
5.a) 20.6 m/s
14 degrees South of East
b) 5.0 seconds
c) 25 meters
VECTOR ADDITION
1) 165 m, 40 degrees South of West
2) 240 m/s, 28 degrees North of East
3) 34 m/s, 33 degrees North of East
4) 10 Newtons, 0 degrees
5.a) 105 Newtons at 93 degrees
b) 105 Newtons at 273 degrees
For some reason, everyone thought we could leave as i left for bathroom!!! Sorry, Mrs. Kozoriz!
DON'T FORGET TO STUDY FOR THE TEST TOMORROW! aaaaannnnddddd... the next scribe is CRAIG!
Tuesday, September 11, 2007
Relative Velocities!
Hello it's me Aichelle, from the third period class...and I am filling in for Vincent today due to the lack of time he has today. So sit back and enjoy [or not].
In today's class, we got out assignments back, which we handed in yesterday. Yay! Okay so after that, we went over questions 21-26 on the sheet we received yesterday. Those questions were pretty straight forward, so there's no reason to go over them. If you need a reference see slide one.
Next, we moved onto Relative velocities in our textbooks found on page ninety five. Ms. K. put the words "Relative Motion" and "Reference frame" onto the smartboard. We then had a little discussion on what each term was. She also asked us to explain it in our own words. It was also in the textbook. Just incase anyone missed it...
Relative motion - the motion of an object or a person in respect to another object/person.
Reference frame (or frame of reference) - the object or system with respect to which velocity is measured.
After a bit of story time from Ms. K., she then moved on to explaining the different cases of Tarzan, etc. [Slide 2]
Case 1
This case dealt with Tarzan. [Slide 3]
In today's class, we got out assignments back, which we handed in yesterday. Yay! Okay so after that, we went over questions 21-26 on the sheet we received yesterday. Those questions were pretty straight forward, so there's no reason to go over them. If you need a reference see slide one.
Next, we moved onto Relative velocities in our textbooks found on page ninety five. Ms. K. put the words "Relative Motion" and "Reference frame" onto the smartboard. We then had a little discussion on what each term was. She also asked us to explain it in our own words. It was also in the textbook. Just incase anyone missed it...
Relative motion - the motion of an object or a person in respect to another object/person.
Reference frame (or frame of reference) - the object or system with respect to which velocity is measured.
After a bit of story time from Ms. K., she then moved on to explaining the different cases of Tarzan, etc. [Slide 2]
Case 1
This case dealt with Tarzan. [Slide 3]
Case 2
This case dealt with Tarzana and was similar to Tarzan's but it had an extra step which involved using the tan function to find the direction [angle]. [Slide 4]
After that, we seemed to get the gist of it. So we went through question three but not in extensive detail. Next we were given an assigment on page 116: 37-42. which can be found on slide 5.
Since I'm feeling generous I will type out all of the questions onto here.
37. A boat wishes to travel east. If there is a current of 10km/h flowing north and the boat is capable of travelling at 30km/h, find the heading and velocity of the boat as seen by a person on the shore.
38. A plane is seen to travel in a direction [S30°W]. If it's ground velocity was 300km/h and the wind speed is 150km/h south, what is the plane's velocity relative to the air?
39. A boat wishes to cross a lake and end up directly south of where it started. The boat is capable of moving 340km/h. If there is a current of 8.0km/h flowing to the west, find
a) the heading the boat must take in order to successfully complete the trip
b) the velocity relative to the ground
40. How much faster is it to point yourself and swim directly across instead of fighting the current and swimming directly across if the width of the river is 100km, you are capable of swimming at 2.2m/s and there is a current of 1.6m/s?
41. A large cruise boat is moving at 15km/h east relative to the water. A person jogging on the ship moes across the ship in a northerly direction at 6km/h. What is the velocity of the jogger relative to the water?
42. A shortstop running at 2.om/s towards third base catches and throws a ball toward homeplate at 35m/s. If the shortstop and catcher are lined up in a direct line at sight when the shortstop throws the ball,
a) at what angle to his body should he throw in order for the ball to move in a straight line directly from him to the catcher?
b) if the distance the ball travels is 20m, how long does the ball take to get to the catcher?
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Just a few reminders:
*sign up for the blog if you haven't already by e-mailing Ms. K.
*use labels in scribe posts.
*try to go beyond your limits while doing scribe posts. [use graphics/different colours/fonts/explanations]
*do your homework.
*Vincent's tomorrow's scribe.
*We will have a test on Thursday.
---hopefully you all got something out of this...=) because that would be super awesome and if I made a mistake, please let me know, so I can fix it toute de suite [right away].
This case dealt with Tarzana and was similar to Tarzan's but it had an extra step which involved using the tan function to find the direction [angle]. [Slide 4]
After that, we seemed to get the gist of it. So we went through question three but not in extensive detail. Next we were given an assigment on page 116: 37-42. which can be found on slide 5.
Since I'm feeling generous I will type out all of the questions onto here.
37. A boat wishes to travel east. If there is a current of 10km/h flowing north and the boat is capable of travelling at 30km/h, find the heading and velocity of the boat as seen by a person on the shore.
38. A plane is seen to travel in a direction [S30°W]. If it's ground velocity was 300km/h and the wind speed is 150km/h south, what is the plane's velocity relative to the air?
39. A boat wishes to cross a lake and end up directly south of where it started. The boat is capable of moving 340km/h. If there is a current of 8.0km/h flowing to the west, find
a) the heading the boat must take in order to successfully complete the trip
b) the velocity relative to the ground
40. How much faster is it to point yourself and swim directly across instead of fighting the current and swimming directly across if the width of the river is 100km, you are capable of swimming at 2.2m/s and there is a current of 1.6m/s?
41. A large cruise boat is moving at 15km/h east relative to the water. A person jogging on the ship moes across the ship in a northerly direction at 6km/h. What is the velocity of the jogger relative to the water?
42. A shortstop running at 2.om/s towards third base catches and throws a ball toward homeplate at 35m/s. If the shortstop and catcher are lined up in a direct line at sight when the shortstop throws the ball,
a) at what angle to his body should he throw in order for the ball to move in a straight line directly from him to the catcher?
b) if the distance the ball travels is 20m, how long does the ball take to get to the catcher?
-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-~-
Just a few reminders:
*sign up for the blog if you haven't already by e-mailing Ms. K.
*use labels in scribe posts.
*try to go beyond your limits while doing scribe posts. [use graphics/different colours/fonts/explanations]
*do your homework.
*Vincent's tomorrow's scribe.
*We will have a test on Thursday.
---hopefully you all got something out of this...=) because that would be super awesome and if I made a mistake, please let me know, so I can fix it toute de suite [right away].
Monday, September 10, 2007
today's class, today's scribe (09/10/07)
Today's class we handed in the Acceleration Problems worksheet for marks.
Also we read pages 95-103 and answered questions 35 and 36 on page 116 in the text book based the topic Relative Motion. Worksheets were also given to us based on the topic... if you need the worksheets ask Ms K for it.
-Russel L.
the next scribe will be.... vincentr
Also we read pages 95-103 and answered questions 35 and 36 on page 116 in the text book based the topic Relative Motion. Worksheets were also given to us based on the topic... if you need the worksheets ask Ms K for it.
-Russel L.
the next scribe will be.... vincentr
Sunday, September 9, 2007
Scribe List
This is The Scribe List. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.
This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.
This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.
St.Jofe | 10109547 Sharmaine D |
09/07/07
Went over both acceleration work sheets (answers below this post). We then went through "Study Guide 4.2 Displacement during constant acceleration". After that we were assigned to do 11 question on a acceleration problem work sheet which will be taken in on Monday September 10 for marks. That's what went down on Friday Sept. 7, see you guys Monday
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Friday, September 7, 2007
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