# physics question help!

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#1
Somebody plz help me on this physics question. i dont undastand it so plz can u post the answer with explanation. i would b so greatful!
here is the question:
an ipod battery in use delivers 4.4w. if the batttery was replaced by a square handheld solar cell, what would the side length of the solar cell need to be? assume all sunlight energy landing on the cell is converted into electricity.

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10 years ago
#2
(Original post by mujahid_e3)
Somebody plz help me on this physics question. i dont undastand it so plz can u post the answer with explanation. i would b so greatful!
here is the question:
an ipod battery in use delivers 4.4w. if the batttery was replaced by a square handheld solar cell, what would the side length of the solar cell need to be? assume all sunlight energy landing on the cell is converted into electricity.

To answer this question, you need to know how much Sun's power reaches each square metre of the Earth's surface (it is called irradiance). From that you can work out how big an area your solar panel needs to be and then the length of one side of it.
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10 years ago
#3
There must be more info than that because you can't work it out from the info you have provided
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#4
(Original post by Mbob)
To answer this question, you need to know how much Sun's power reaches each square metre of the Earth's surface (it is called irradiance). From that you can work out how big an area your solar panel needs to be and then the length of one side of it.
so how do you go about solving it?? I havent been provided that information?
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#5
(Original post by dentistry1)
There must be more info than that because you can't work it out from the info you have provided
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#6
(Original post by DontJudge)
you have to put it into that equation. !!! you need to know them by heart!
Thanks
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10 years ago
#7
Solar panels -> Ipod is being powered by the sun -> Need to know the energy per unit area reaching the Earth, from the sun. -> Unsolvable otherwise. As said before.
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10 years ago
#8
(Original post by mujahid_e3)
You CAN NOT work it out from that info.
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10 years ago
#9
Is this an exam question or is it a teaser question (ie supposed to be very difficult)?
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10 years ago
#10
(Original post by dentistry1)
There must be more info than that because you can't work it out from the info you have provided
Lmao...

You've been told everything. All you have to do is look up in your book or on the internet or wherever how much energy hits the surface of the earth from the sun.
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10 years ago
#11
(Original post by insoms)
Lmao...

You've been told everything. All you have to do is look up in your book or on the internet or wherever how much energy hits the surface of the earth from the sun.
No because then you would have to know longitude/latitude and the time and date.
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10 years ago
#12
(Original post by dentistry1)
No because then you would have to know longitude/latitude and the time and date.
I don't think its that complicated a question you might be overcomplicating it abit.

Using that level of depth you'd have to specify the exact location on Earth for every Mechanics question, because the value of g changes minutely wherever you are. No we just take g as 9.81 and move on with it .
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10 years ago
#13
Yeah, definitely agreeing that there should be an average value somewhere. I refuse to believe you do not either have a formula, or access to a value for this.

_Kar.
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#14
(Original post by dentistry1)
You CAN NOT work it out from that info.
Ok if i have the values what formula do i use to work it out??
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#15
(Original post by insoms)
Lmao...

You've been told everything. All you have to do is look up in your book or on the internet or wherever how much energy hits the surface of the earth from the sun.
Ok after i find how much energy hits the surface of the earth from the sun what formula do i use? Please help

Thanks
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10 years ago
#16
(Original post by insoms)
I don't think its that complicated a question you might be overcomplicating it abit.

Using that level of depth you'd have to specify the exact location on Earth for every Mechanics question, because the value of g changes minutely wherever you are. No we just take g as 9.81 and move on with it .
g only differs slightly, irradiance differs massively depending on longitude/latitude/time/date so it absolutely has to be taken into account.
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10 years ago
#17
(Original post by Kareir)
Yeah, definitely agreeing that there should be an average value somewhere. I refuse to believe you do not either have a formula, or access to a value for this._Kar.
I agree with this^ otherwise it if far too complex for school physics
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10 years ago
#18
(Original post by mujahid_e3)
Ok after i find how much energy hits the surface of the earth from the sun what formula do i use? Please help

Thanks
Physics isn’t about plugging numbers into equations. You can answer this questions without needing to “know “ any equations at all. If you know the Sun’s irradiance (power per square metre) and know how much power you need to run the IPod, then you can work out the area of the solar panel you need (in square metres). Then, given that it is a square, you can work out the length of a side.

If you’re not given the irradiance of the Sun (I'm surprised you weren't, do you have a data sheet), google it and take a typical figure (it won't be exact because it depends on location etc.). For example:

http://en.wikipedia.org/wiki/Sunligh....29_upon_Earth

The point of these type of questions isn't to get the right answer, it's to show you understand the basic physics behind it.

I can’t give you any more help without giving you the answer, which would benefit no-one!
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#19
(Original post by Mbob)
Physics isn’t about plugging numbers into equations. You can answer this questions without needing to “know “ any equations at all. If you know the Sun’s irradiance (power per square metre) and know how much power you need to run the IPod, then you can work out the area of the solar panel you need (in square metres). Then, given that it is a square, you can work out the length of a side.

If you’re not given the irradiance of the Sun (I'm surprised you weren't, do you have a data sheet), google it and take a typical figure (it won't be exact because it depends on location etc.). For example:

http://en.wikipedia.org/wiki/Sunligh....29_upon_Earth

The point of these type of questions isn't to get the right answer, it's to show you understand the basic physics behind it.

I can’t give you any more help without giving you the answer, which would benefit no-one!

Ok if you can just post the answer and i will see your working out and see if i understand

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10 years ago
#20
(Original post by mujahid_e3)
Ok if you can just post the answer and i will see your working out and see if i understand

If, for example, the amount of incident power from the sun is 2.2w/m^2,

You would need 2 square metres of solar panel to catch 4.4 watts, which would power the iPod.

_Kar.

NB: It really, really isn't 2.2w/m^2.
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