Sorry for the late post, for thsoe who check this blog regularly, I kinda forgot and just remebered this morning. Well Yesterday we didnt do much. We learned about the weight and scale and how it REALLY feels when you move it up or down. We were also given a Lab to do which is probably due today if anyone is unsure.
Heres the notes she gave us, please make sure to correct me if I'm wrong because I'm writing this based on memory.Also sorry that I can't post any diagrams to explain this better.
F=mg
At this point the weight is at rest so it will feel normal.
F=mg + ma
You move the weight and scale along with it up and this is how it will feel...
It will become heavier at first then at a constant velocity you will feel normal. Then as you start to slow down as you reach the end the weight then becomes lighter.
F=mg -ma
This one is opposite direction with the opposite affect. You feel light at first as you move the weight and scale in the downward direction, then at constant velocity it feels normal. When you the start to slow down towards the end it will then become heavy.
*NOTE* When I say it feels normal it just means you can't tell if it got heavier or lighter then before.
The Lab...
For those who were confused on what they needed to do
What we had to do was get a rubber thingy(don't rember what its called) and tie to to a piece of string that is about 1.5 m's long. You then shorten the string by rolling in your hand to get 1.2m, 1.0m, 0.8m, 0.6m, 0.4m ,0.2m. These are the different lengths that you will be measuring how much time its takes for it to swing twenty cycles( one cycle equals the time it takes for it to come back from where you started it. Remember to start the rubber stopper up to the right of you and keep hand still will doing experiment. Its quite annoying to hold all the time so switch places with your partner if your arm feel tired. Record all the data 3 times each. Then average your time. The period is calculated by the (average time/20 cycles it did). Of course do the Analysis questions after you have completed this experiment and dont forget to hand in.
Sorry once again for the late post as I COMPLETELY FORGOT even though I was told before hand. :S
The next scribe I choose will be Alex Sharp. So good luck Alex with next scribe.
Showing posts with label Anthony Lee. Show all posts
Showing posts with label Anthony Lee. Show all posts
Wednesday, December 17, 2008
Tuesday, November 25, 2008
Monday & Tuesday, November 24-25,2008
Today I'll be doing two blogs since I forgot to do it for monday and assign someone for Tuesday. First we corrected alot of sheets in physics, "Components of vectors", first page of Acceleration, 4.1 What is acceleration?
Components of vectors
1. b
2. b
3. c
4. b
5. d
6. a
7. c
8. c
9. b
10. c
11. c
12. c
13. c
14. c
Acceleration (first page)
1. a = V2-V1/Change of time = 25m/s-0m/s / 4s-0s = 6.3 m/s²
2. a = V2-V1/change of time = 29.28m/s-0m/s / 3s-0s = 9.8 m/s²
3. V2 = V1 + a(change of time) = 28m/s + (2.5m/s²)(3s-0s) = 36m/s
4. change of time = V2-V1 / a = 14.1m/s -0m/s / 13.2m/s² = 4.4s
5. change of velocity = a(change of time) = 56.3m/s²(1.9) = 107m/s = 110m/s with sig. digits
What is Acceleration?
The change in velocity divided by the time interval is average acceleration. It can be calculated using the equation a=v/t. In this equation a stands for acceleration, ٨V stands for change in velocity, and ٨t stands forthe time interval. If velocity is measured in metres per second, acceleration is measures in m/s/s, which is read as metres per second per second. The unit also can be written as m/s², which is read as metres per second squared. Like velocity, acceleration is a(n) vector quantity, which means it has both magnitude and direction. When velocity increases, acceleration is positive. When velocity decreases, acceleratin is negative.
Average and Instantaneous Acceleration
A velocity-time graph shows how velocity depends on time. The rise of the curve represents the change in velocity. The run of the curve represents the time interval. The slope of the curve represents the average acceleration. If the curve on a velocity-time graph is a straight line, the acceleration is constant. If the curve is not a straight line, acceleration is changing. The slope of a line tangeant to the curve is the instantaneous acceleration at that time.
Velocity of an object with constant acceleration
Acceleration that does not change in time is constant, or uniform, acceleration. The velocity when the clock time is zero is the initial velocity. The velocity after acceleration has occurred is called the final velocity, and is calculated using the equation V2=V1+at. In this equation, V2 is final velocity, V1 initial velocity, a is acceleration, and t is time interval.
Displacement when velocity and time are known
If an object is accelrating, its displacement can be calculated using the equation d=(V2+V1/2)t. In this equation, d stands for displacement, V2 stands for final velocity, V1 stands for initial velocity, and ٨t stands for time interval. To find displacement using a velocity-time graph, find the area under the curve.
Well this is all i can remember correcting over these two days, tell me if i forgot something and i'll try to find the answers to those also. Also don't forget to do the 2nd page on "Acceleration" as we will probably correct it in class tomorrow.
TOMORROW'S SCRIBE : Suzette
Components of vectors
1. b
2. b
3. c
4. b
5. d
6. a
7. c
8. c
9. b
10. c
11. c
12. c
13. c
14. c
Acceleration (first page)
1. a = V2-V1/Change of time = 25m/s-0m/s / 4s-0s = 6.3 m/s²
2. a = V2-V1/change of time = 29.28m/s-0m/s / 3s-0s = 9.8 m/s²
3. V2 = V1 + a(change of time) = 28m/s + (2.5m/s²)(3s-0s) = 36m/s
4. change of time = V2-V1 / a = 14.1m/s -0m/s / 13.2m/s² = 4.4s
5. change of velocity = a(change of time) = 56.3m/s²(1.9) = 107m/s = 110m/s with sig. digits
What is Acceleration?
The change in velocity divided by the time interval is average acceleration. It can be calculated using the equation a=v/t. In this equation a stands for acceleration, ٨V stands for change in velocity, and ٨t stands forthe time interval. If velocity is measured in metres per second, acceleration is measures in m/s/s, which is read as metres per second per second. The unit also can be written as m/s², which is read as metres per second squared. Like velocity, acceleration is a(n) vector quantity, which means it has both magnitude and direction. When velocity increases, acceleration is positive. When velocity decreases, acceleratin is negative.
Average and Instantaneous Acceleration
A velocity-time graph shows how velocity depends on time. The rise of the curve represents the change in velocity. The run of the curve represents the time interval. The slope of the curve represents the average acceleration. If the curve on a velocity-time graph is a straight line, the acceleration is constant. If the curve is not a straight line, acceleration is changing. The slope of a line tangeant to the curve is the instantaneous acceleration at that time.
Velocity of an object with constant acceleration
Acceleration that does not change in time is constant, or uniform, acceleration. The velocity when the clock time is zero is the initial velocity. The velocity after acceleration has occurred is called the final velocity, and is calculated using the equation V2=V1+at. In this equation, V2 is final velocity, V1 initial velocity, a is acceleration, and t is time interval.
Displacement when velocity and time are known
If an object is accelrating, its displacement can be calculated using the equation d=(V2+V1/2)t. In this equation, d stands for displacement, V2 stands for final velocity, V1 stands for initial velocity, and ٨t stands for time interval. To find displacement using a velocity-time graph, find the area under the curve.
Well this is all i can remember correcting over these two days, tell me if i forgot something and i'll try to find the answers to those also. Also don't forget to do the 2nd page on "Acceleration" as we will probably correct it in class tomorrow.
TOMORROW'S SCRIBE : Suzette
Tuesday, September 30, 2008
Tuesday, September 30, 2008
Hey Everyone! I'm Anthony Lee and I'll be your scribe today! There are two Anthony's in our class so please dont get me confused with the other one :). Since we didn't do much except correct questions 5-13 on the back of our booklet "WAVES IN TWO DIMENSIONS." So this will be short and sweet.
ANSWERS TO Q's 5-13 on WAVES IN TWO DIMENSIONS
5.
This diagram shows that when a water wave hits a barrier, the one with the bigger wavelength will diffract more. So to answer question 5 the answer is "Situation B" shows the greatest diffraction.
6. The ocean wave with a 200m wavelength
7.. THe D note (294 Hz) will be heard more clearly because the frequency is lower therefore having a bigger wavelength.
7.. THe D note (294 Hz) will be heard more clearly because the frequency is lower therefore having a bigger wavelength.
8. The orange light (610nm)
9.
a) Red light (700 nm)
b) Red light (700 nm)
10. 400m wave crests
11.
a) 10000Hz
b) 10000Hz
12.
a) The 2nd nodal to the left of prependicular bisector is the orange line
b) It is destructive interence at the point of the square
c) square is located on nodal line 4
d) (4.0)(6 wavelengths)= 24 cm
13.
a) (16 cm - 15 cm) = (1-1/2) ﮑ
1 cm = 1/2 ﮑ
2 cm = ﮑ
1 cm = 1/2 ﮑ
2 cm = ﮑ
b) V = f ﮑ
= (4.0 Hz)(2 cm)
= 8 cm/s
*Reminder* For those who are having trouble with diffraction just remember that when wave length is wider the diffraction will be greater, and if frequency is lower the diffraction will be greater.
Well this is about all we did, except we were given a review but that will be corrected tomorrow so we can let the next scribe worry about it. AND THE PERSON I CHOOSE FOR NEXT SCRIBE IS ALEX SHARP! cya all in class tomorrow!
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