asib1214
  • asib1214
Yelp!!!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rvc
  • rvc
@Michele_Laino
rvc
  • rvc
what is your ans?
Michele_Laino
  • Michele_Laino
since we have three complete cycles within 4.8 meters, then the wavelength is: 4.8/3=1.6 meters

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rvc
  • rvc
since we have 3 waves so divide 4.8 by 3 because one wave is equal to wavelength
rvc
  • rvc
lol sorry i did not see your post @Michele_Laino
Michele_Laino
  • Michele_Laino
no worries :) @rvc
Michele_Laino
  • Michele_Laino
the requested velocity, is: 4.8/6=...meters/seconds
Michele_Laino
  • Michele_Laino
period is T= 6 /3 =...second frequency f= 1/T =... Hz
Michele_Laino
  • Michele_Laino
if the velocity of our wave is unchanged, then we can apply this formula: \[\Large \lambda f = v\] from which we get: \[\Large \lambda = \frac{v}{f} = \frac{{0.8}}{2} = ...meters\]
asib1214
  • asib1214
this is how i did it, i first took the inverse of the Ferquency.....F=1/T which agve me the period
Michele_Laino
  • Michele_Laino
how did you get 14.5 meters/second
Michele_Laino
  • Michele_Laino
v=traveled space/time interval= 4.8/6=0.8 m/second
Michele_Laino
  • Michele_Laino
period is T=6/3= 2 seconds
Michele_Laino
  • Michele_Laino
I think taht: since within 6 seconds we have three complete waves, then the period is:T=6/3= 2 seconds
Michele_Laino
  • Michele_Laino
so the corresponding frequency is: f=1/T= 1/2=0.5 Hertz
asib1214
  • asib1214
when solving for the very first question wavelength.
asib1214
  • asib1214
i mean second cuz time is given after the wavelength question :(
asib1214
  • asib1214
so do we assume we don't know the time while solving for wavelength
Michele_Laino
  • Michele_Laino
as I wrote before, in the last part of the first question, we have to assume that the speed of our wave is the same as before, namely, it is 0.8 m/second
asib1214
  • asib1214
@rvc do i figure out the amplitude? Please help!!!
asib1214
  • asib1214
@Astrophysics
asib1214
  • asib1214
how do i figure out the amplitude? Please help!!!
asib1214
  • asib1214
the picture is at the top.
Astrophysics
  • Astrophysics
|dw:1435727528455:dw| you can find the amplitude using \[Amplitude = \frac{ Distance }{ Frequency }\]
Astrophysics
  • Astrophysics
They are related.
Astrophysics
  • Astrophysics
No, the amplitude here is measured in decibels here. I think you're just confusing it because of the formula/ units but not really sure about what this actually means.
Astrophysics
  • Astrophysics
\[\lambda = \frac{ v }{ f }\]
asib1214
  • asib1214
can you quickly guide me what to do Please?!?!?!?!
Astrophysics
  • Astrophysics
Slow down, ok so we have |dw:1435728513829:dw| so we have 3 complete cycles here, so our wavelength is 4.8/3
Astrophysics
  • Astrophysics
\[y = A \sin \omega t\] you can use this A here is the amplitude the omega is angular frequency and t is the time period.
Astrophysics
  • Astrophysics
\[\omega = f 2 \pi\]
Astrophysics
  • Astrophysics
It's one in the same, math and physics are interrelated...
Astrophysics
  • Astrophysics
Lol, no problem, it can get confusing I guess, but I think you just need a decent understanding of what everything means and you will get it, you can try khan academy they have some nice videos on such topics I'm sure
rvc
  • rvc
i m extremely sorry @asib1214 i was away from the laptop for a long time. Thanks @Astrophysics :) Sorry once again :)
rvc
  • rvc
Do you have the formula for amplitude?
rvc
  • rvc
so use the formula :)
rvc
  • rvc
The maximum displacement of the particle from the mean position is the amplitude.
rvc
  • rvc
:/
rvc
  • rvc
v=n lambda
rvc
  • rvc
:)
rvc
  • rvc
idk :(
rvc
  • rvc
@welshfella please help us
welshfella
  • welshfella
i'll have to look later if you still need it gotta go right now
welshfella
  • welshfella
there are 3 complete waves in the diagram so one wavelength is 4.8 / 3 m
welshfella
  • welshfella
well the amplitude is the distance from the equilibrium point to the crest of the wave. I'm not sure if we have enough information ..
welshfella
  • welshfella
i'll have to check that out.
welshfella
  • welshfella
A = D/F where D is distance travelled by the wave / frequency F
welshfella
  • welshfella
I'm a bit rusty with this stuff.
welshfella
  • welshfella
yes i see hmmm
welshfella
  • welshfella
1.6 m is correct for the wavelength i'm sure of that
welshfella
  • welshfella
yea sorry my recall of this stuff is really hazy. one website i visited gave the formula A = D / F so i guess its must be right
welshfella
  • welshfella
D is distance and F = freqency in cycles per second
welshfella
  • welshfella
yea sorry i cant be of more help. If i have the time i'll have to revise this stuff - I did it in Physics years back!!
Michele_Laino
  • Michele_Laino
hint: in order to find the amplitude, you have to measure the distances D and H as below:
1 Attachment
Michele_Laino
  • Michele_Laino
you have to measures the distances H adn D on your drawing, using a scale ruler, then you have to solve this proportion: \[\Large D:4.8 = H:x\] where x is the requested amplitude
Michele_Laino
  • Michele_Laino
so we have: \[\Large x = \frac{{4.8 \times H}}{D} = ...\]
Michele_Laino
  • Michele_Laino
what is the distance D? Please measure it with a scale ruler
Michele_Laino
  • Michele_Laino
how many cm is D? how many cm is H?
Michele_Laino
  • Michele_Laino
ok! and how many cm is D?
Michele_Laino
  • Michele_Laino
ok! then if we apply our formula, we get: \[\Large x = \frac{{4.8 \times H}}{D} = \frac{{4.8 \times 1}}{{10.5}} = ...meters\]
Michele_Laino
  • Michele_Laino
what is x?
Michele_Laino
  • Michele_Laino
That's right! better is x=0.457 meters
Michele_Laino
  • Michele_Laino
it is a proportion
Michele_Laino
  • Michele_Laino
we have finished, since by means of that proportion, our amplitude, is 0.457 meters
Michele_Laino
  • Michele_Laino
yes! I am sure.
Michele_Laino
  • Michele_Laino
yes! since our procedure is correct!
Michele_Laino
  • Michele_Laino
yes! that's right!
Michele_Laino
  • Michele_Laino
thanks!
rvc
  • rvc
Thanks @Michele_Laino
Michele_Laino
  • Michele_Laino
:) @rvc
asib1214
  • asib1214
@Astrophysics A tuning fork with a frequency of 420Hz emits sound with a wavelength of 0.82m in air. If the temperature of the air increases, what will happen to the wavelength and why?
Astrophysics
  • Astrophysics
http://www.sengpielaudio.com/calculator-speedsound.htm
asib1214
  • asib1214
i understand that the rectangular wave on the left has a positive displacement because it's above the equilibrium and the triangle trough has a negative displacement, the thing that i don't understand here is that, Why Triangle trough DOESN'T have any displacement?!?!?!?!
Michele_Laino
  • Michele_Laino
1) we have a destructive interference when 2 signals overlap each other and they cancel out each other, namely during that overlapping we have no signal, for example the dark fringes in a interference pattern 2) we have a constructive interference, when 2 signals overlap each other and they reinforce each other, so during that overlapping we have a signal whose intensity is greater than each of both overlapped signal. For example the bright fringes in a interference pattern
asib1214
  • asib1214
i understand that when two waves interfere with each other, they form a constructive wave, and when two waves with opposite displacement destruct each other!!!
Michele_Laino
  • Michele_Laino
a destructive interference occurs, when from the overlapping between 2 signals, the resultant signal has a little intensity, less than the intensities of each signal
asib1214
  • asib1214
When these weird shapes confuse me, like when a triangle interferes with a rectangle :( like how do you figure out if they are going to form a constructive or destructive waves :(
Michele_Laino
  • Michele_Laino
since the phase shift between the rectangle signal and the triangle signal is 180 degree, namely those signals interfere, each other, with opposite phases
Michele_Laino
  • Michele_Laino
you have to subtract the y-coordinates of the triangle signal, from the corresponding y-coordinate of the rectangle signal
asib1214
  • asib1214
oh so since the triangle is flipped to 180 degress, but still how do i know the resultant displacement or what shape will wave form during the collision??!!?1
Michele_Laino
  • Michele_Laino
as I wrote before, you can get that shape if you subtract the y-coordinates of the triangle signal, from the corresponding y-coordinate of the rectangle signal
asib1214
  • asib1214
can you please draw a quick drawing?!?!!?!?? :(
asib1214
  • asib1214
please!?!?!?!?!?!!?!?
Michele_Laino
  • Michele_Laino
|dw:1436132812859:dw|
Michele_Laino
  • Michele_Laino
|dw:1436132987610:dw|
asib1214
  • asib1214
i'll do the next two question after this example, please check them for me, I Love You!!! :")
Michele_Laino
  • Michele_Laino
then we have: |dw:1436133042674:dw|
asib1214
  • asib1214
OMG i'm crying here, Thank you so much, please stick around, i'll do the next two questions on my own. Please check them for me!!!! Thank you!!!!
asib1214
  • asib1214
wait one quick question here. how do you know the amplitude is 5, can you just assume and make up your own numbers to figureout the resultant displacement.
asib1214
  • asib1214
i nother words, how do yo know what shape is bigger or smaller :(
Michele_Laino
  • Michele_Laino
the amplitude A is given by this computation: \[A = \frac{{\max value - \min value}}{2}\]
asib1214
  • asib1214
OMG i'm going to ask you a very sily question but how did you come up with 5?!?!?
Michele_Laino
  • Michele_Laino
I chose those value arbitrarily
Michele_Laino
  • Michele_Laino
values*
asib1214
  • asib1214
because the picture didn't give any numbers, so when i'm solving questions on my own, how do i assume the amplitude is 5 and 8 :(
asib1214
  • asib1214
oh okay, let me try the next two questions and i'll get back to you.....i hope you'll be here!!! :")
Michele_Laino
  • Michele_Laino
no, you have to measure your numbers using a scale ruler
asib1214
  • asib1214
ok so in what units???? cm?
Michele_Laino
  • Michele_Laino
yes!
asib1214
  • asib1214
ok Thanks you Soooooooooooooooooooooooooooooooooooooooooooooooo MCH!!!! :")
asib1214
  • asib1214
Love you forever!!! :")
Michele_Laino
  • Michele_Laino
Thanks!

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