The document discusses the speed of sound and concepts related to sound waves. It provides examples of calculating the speed of sound given temperature, distance, and time. It also discusses wavelength, frequency, and solving related problems involving sound traveling through different mediums like air and wood. Formulas for speed of sound and related concepts are provided and applied to various word problems throughout the document.

1. Today we went over the following worksheets: :Concept –development practice page :unit 1.3 sound Hi! I’m KIMBERLY n” I’m d’ scribe 4 2day……:>

2. GRADE 11 PHYSICS ;UNIT 1.3 SOUND Answer the following questions in the space provided.The following formula may be helpful in solving this question. V=332m/s+(0.6m/s /o C )( t o C) 1.Given a summer temperature of 20 o C,find the speed of sound. Answer: V=332 m/s+(o.6m/s/ 0 C)(20 0 C) V=332m/s+(0.6m/s /o C )( t o C) 0.6x20=12 332+12= 344m/s Using this formula …plug in the given numbers . Multiply the 0.6 and 20 0 Then,add the product to 332 Don’t forget the unit..

3. 2.Calculate the speed of sound in air and in wood given that it travels 800m and takes 2.35s in air and 0.20 in wood .how many times faster is it in wood? Answers: Air V=d/t V=? D=800m T=2.35s V=800/2.35 =340.43 m/s Wood V=d/t V=? D=800 T=0.20 V=800/0.20 V= 4000 m/s =4000/340.45 = 11.7or 12 The speed of sound is 12 times faster in wood than in air

4. The sound of thunder occurs 12.3 after lightning is seen.how far away is the storm if the temperature is: a.0 o C b.10 o C c.30 o C d.-10 o C V=332m/s S=12.3 V=d/t D=vt 332m/s(12.3) D= 4083.6 Same method of solving as letter A ,but you need to find the velocity first .. using D=4157.2m d=4305m d=4009.8m V=332m/s+(0.6m/s /o C )( t o C)

5. Answer: F=90khz---90,000hz (convert it first..) Temp-22 o C Wavelength=? V=322m/s+(0.6m/s/ o C)(t o C) V=322m/s+(0.6m/s/ o C)(22 o C) 332m/s+13.2 =345.2m/s To find the wavelength….use the info …above Wavelength =v/f =345.2m/s/90,000hz =.00383m Answer: Convert the 5300km/h----1472 m/s 870millis-----0.8705s Then solve it using the formula…. d=vt =(1472m/s)(0.4355) =640m Half of 0.8705s..

6. Answer: D=272m T=0.8s Temp.=? Use the v=d/t to find the velocity V=272m/0.8 V=340 Subtract the 340m/s to 332m/s 340m/s-332m/s=8 Then…divide the 8 to 0.6 =8/0.6 =13.3 o C V=332m/s+(0.6m/s /o C )( t o C) V=332m/s+(0.6m/s /o C)(25 o C) =332m/s+15 =347m/s Same method of solving as letter A , =323m/s

7. Answer: V=d/t 344m/s=d/1.0 D=344m/s Half of 2.0s V=332m/s+(0.6m/s /o C )( t o C

8. longitudinal Pitch 2.5

9. long behind 9 seconds

10. solid 4 sec.. Will not

11. 220 hz Assume that the room temp..is 20 o C V=332m/s+(0.6m/s /o C )( t o C) V=332m/s+(0.6m/s /o C )( 20 o C) 332m/s+12 = 344m/s

12. Don’t forget to answer our homework…. Booklet :… #10 page 27 #14,16,20 page 28 And the sound waves worksheet