Wednesday, October 15, 2008

October 15, 2008

Hi guys! This is Jeamille and I'll be your scribe for today...

First off, Ms. K. gave us back our lab sheet from last week called Sounds and Their Sources and the sheet from yesterday (10 points about the video). After this, we corrected the papers that we were supposed to finish from Thursday. These sheets were called Sound Waves, Unit 1.3 Sound, Concept-Development Practice Page 26-1 and Career of the Month (we just had to read this one).

Here are the corrections:

Concept-Development Practice Page 26-1
1) Longitudinal
2) Pitch
3) 2.5 cm
4) Long
5) a. Behind b. 9 seconds
6) Solids
7) 4 seconds
8) Will not -- Every other push will oppose the motion of the swing
9) 220 Hz
10) Slightly different frequencies
11) 4 Hz
12) 346 m/s

Unit 1.3 Sound
1) 344 m/s
2) 11.75 faster in wood
3) a. 4 083.6 m b. 4 150 m c. 4 305 m d. 4 009.8 m
4) 0.00385 m
5) 640.3 m
6) 13.3 °C
7) a. 347 m/s b. 320 m/s
8) 344.6 m

Sound Waves
1) 1 218 m
2) 8.77 s
3) 1.19 m
4) 642 m
5) v = 330 m/s + (20 C)(.6) = 342 m/s
The echo from 1 wall returns in 0.5 seconds so you take 1/2 the time which is 0.25 s and multiply it by 342 m/s to find the distance to that wall. 342 m/s * .25 s = 85.5 m

Then the time it took for the other echo to return to the person was 0.5 s after he heard the first echo (which was 0.5 s). The total time was 0.5 s + 0.5 s = 1.0 s. You take 1/2 of this time which is 1/2 * 1 s = 0.5 s
The distance to this wall is the speed * time = 342 m/s * 0.5 s = 171 m.
Add the two distances together, 85.5 m + 171 m = 256.5 m.

(this is from Ms. Kozoriz :D)

Before we went on #5 in Sound Waves, Ms. K. gave us a sheet called Consonance and Frequency and showed us how she got the answer for that one.

After correcting these, we did a lab. :) It was pretty easy, I should say. All we needed were straws and scissors. We had to cut the end of the straw creating an almost triangle shape and blow it to produce sound. While observing what happens, we had to answer 2 sheets for this lab and hand it in by the end of class. For those who didn't finish, Ms. K. said it's okay to hand it in tomorrow.

Well that's it for today's class. :))

Next scribe is Charizze a.k.a Chae

1 comment:

Ms K said...

5. v = 330 m/s + (20 C)(.6) = 342 m/s
The echo from 1 wall returns in 0.5 seconds so you take 1/2 the time which is 0.25 s and multiply it by 342 m/s to find the distance to that wall. 342 m/s * .25 s = 85.5 m

Then the time it took for the other echo to return to the person was 0.5 s after he heard the first echo (which was 0.5 s). The total time was 0.5 s + 0.5 s = 1.0 s. You take 1/2 of this time which is 1/2 * 1 s = 0.5 s
The distance to this wall is the speed * time = 342 m/s * 0.5 s = 171 m.
Add the two distances together, 85.5 m + 171 m = 256.5 m.