Question

An ambulance with its siren wailing is approaching an observer. Write two or three sentences to compare the pitch of the sound for the observer with the pitch for the ambulance driver. What is the term used to describe this phenomenon?

Answer 1

well, for a sound wave that is moving with respect to the observer, it gets its pitch distorted based on the compression of sound waves, do you remember what this phenomenon is called? starts with a d

Answer 2

well the decibal scale is used to measure different intensities, that isn't it is it?

Answer 3

the name is the Doppler Effect

notice how the soundwaves become more compressed as the ambulance approaches the observer, raising the pitch

any ideas how this would compare to the person driving the ambulance? is the siren distorted at all for the driver?

Answer 4

well, since the driver is moving at the same speed as the siren, the driver hears the sound without any pitch distortion

Answer 5

The driver is moving at the same speed as the siren, as a result of this, the driver does not have any pitch distortion. This phonomenon is called the Doppler Effect.

Answer 6

Sorry I was AFK

Answer 7

well good but you also need to include how the sound is affected for the bystander

Answer 8

The sound is affected by the bystander as well. Because of this sound, it is not distorted

Answer 9

is this right?

Answer 10

:/

Answer 11

like it would stay the same for the driver

Answer 12

and for the bystander the closer the driver is getting the higher th epitch

Answer 13

so if you scroll up a bit, we said that for the bystander, as the ambulance drives by the waves get compressed and the pitch goes up

yeah that

Answer 14

So is this good for the final (in my own words):

The driver is moving at the same speed as the siren, as a result of this, the driver does not have any pitch distortion. This phonomenon is called the Doppler Effect. Also, for the bystander as the ambulance drives by him or comes closer to him the pitch goes up and it gets louder for him.

Answer 15

good but it would be a good idea to move the sentence with the doppler effect at the end

Answer 16

* to the end

Answer 17

A light wave travels through water (n = 1.33) at an angle of 28°. What angle does it have when it passes from the water into glass (n = 1.5)?

Answer 18

the water is the original medium so for n1 and theta1 we use the values for water

for n2 we use glass

plug everything in and solve for theta2

Answer 19

(ignore the c parts they're not relevant here)

Answer 20

1.5=1.33/1.5=sin(28)/Sin(x)

x=2.3

Answer 21

hm. I got x = 24.6 degrees

your calculator might be in radians mode

Answer 22

. A sound is transmitted through equal lengths of air and steel. In which material does sound travel faster? Write one or two sentences to explain your reasoning.

Answer 23

air I think?

Answer 24

And because it is not a solid and has an open surface to move faster?

Answer 25

generally the order of sound speeds is

solids > liquids > gases

sounds move faster in solids because the particles are closer together, allowing them to transmit waves more easily

Answer 26

Between these two examples, the solid item would move the fastest, which is steel. This is so because in a solid the molecules are the closest in comparison to a liquid or gas. This allows there to be more wave movements.

Answer 27

A+

Answer 28

Should I get maybe @rishavraj to help with this one

Answer 29

or i think there was a chart maybe

Answer 30

http://www.physicsclassroom.com/class/refln/Lesson-3/Ray-Diagrams-Concave-Mirrors

well I can try to draw it from what I remember

Answer 31

@zarkam21 try this
you will be able to do it by yourself :)
http://www.physicsclassroom.com/class/refln/Lesson-3/Ray-Diagrams-Concave-Mirrors

Answer 32

oops @Vocaloid already posted it ... :)
@zarkam21 lemme know if u need further help :)

Answer 33

oh well yeah I guess you could try the steps, we can try helping if you're getting stuck

Answer 34

next would be drawing how that top ray bounces back from the lens and crosses the focal point, would you mind trying this step?

Answer 35

would be this right

Answer 36

sort of, just be careful because your image is in a different location/size than that diagram

the top of the image is where your rays should converge

Answer 37

@zarkam21 nah that incorrect ... for ur question the object is beyond Center of Curvature..
but the new image u posted.. object is in mid of R and 2R

Answer 38

can you try drawing a straight line from this point, crossing the focal point?

Answer 39

@zarkam21 now make another ray which passes thru the F and then gets reflected back to travel parallel to the principal axis

Answer 40

good but this needs to be extended just a bit so it touches the image

Answer 41

then we start from the top of the object (the black arrow), draw a ray down to the focal point, then touch the mirror, can you try this?

Answer 42

@Vocaloid great job man :)

Answer 43

@rishavraj isnt she just amazing

Answer 44

hm, not quite right, the second ray should point from the object to the focal point

Answer 45

then it bounces off, parallel to the focal axis, like so:

Answer 46

@zarkam21 sorry the orientation is different

Answer 47

now, if we look at the tip of the image, that's where the lines converge so I believe this is done

Answer 48

damn was my reply just ignored ?? o_O *eye roll*

Answer 49

An object is located 85.0 cm from a convex lens. The focal length is 25.0 cm. What is the image distance? Is the image real or virtual?

Answer 50

thank you rishi ;_;

anyway I might let rishi take over b/c I forgot this ;_;

Answer 51

Sure

Answer 52

@rishavraj what do i do here

Answer 53

lemme ask straight.. u want the answer or u wanna understand? :\

Answer 54

I want to understand. I am never here for the answers

Answer 55

@zarkam21 you are aware of the lens formula ?

Answer 56

I forgot that formula? Is it di=do*f

Answer 57

its
\[\frac{ 1 }{ v } - \frac{ 1 }{ u } = \frac{ 1 }{ f }\]

Answer 58

oh okay so it would be

1/85-1/u=1/25

Answer 59

u -> object distance
v -> image distance

Answer 60

1/v-1/85=1/25

v=425/22
v=19.32

Answer 61

so it would be
\[\frac{ 1 }{ v } - \frac{ 1 }{ -85 } = \frac{ 1 }{ 25 }\]

Answer 62

it would be -85
\[\frac{ 1 }{ v } - (\frac{ 1 }{ -85 }) =\frac{ 1 }{ 25 }\]

Answer 63

@zarkam21 done?

Answer 64

OH okay

Answer 65

1/v-1/-85=1/25

v=-425/22
v=-19.32

Answer 66

like this right

Answer 67

@zarkam21 great :)

Answer 68

can we move to the next question

Answer 69

the image would be virtual ?

Answer 70

yes

Answer 71

i thought it was real

Answer 72

@zarkam21 hold on

Answer 73

okay

Answer 74

@zarkam21 solve it again answer would be 35.4 something

Answer 75

1/v-1/-85=1/25

v=-425/12
v=31.4

Answer 76

425/12 = 35.4
-_-

Answer 77

oh im sorry that was a typing mistke

Answer 78

this would be a real image ?

Answer 79

real..

Answer 80

Yayyy

Answer 81

thank you

Answer 82

Can we do another question? Can I tag you

Answer 83

would u not tag me if i said NO :P lol

Answer 84

Lol i eill tage you then