# physics question help!

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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.

please help!

thanks in advance!

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.

please help!

thanks in advance!

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#2

(Original post by

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.

please help!

thanks in advance!

**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.

please help!

thanks in advance!

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#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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(Original post by

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.

**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.

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(Original post by

There must be more info than that because you can't work it out from the info you have provided

**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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(Original post by

you have to put it into that equation. !!! you need to know them by heart!

**DontJudge**)you have to put it into that equation. !!! you need to know them by heart!

Thanks

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#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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#8

(Original post by

Thats the question!! please help havent been provided with anything else!

**mujahid_e3**)Thats the question!! please help havent been provided with anything else!

**work it out from that info.**

__CAN NOT__
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#9

Is this an exam question or is it a teaser question (ie supposed to be very difficult)?

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#10

**dentistry1**)

There must be more info than that because you can't work it out from the info you have provided

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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#11

(Original post by

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.

**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.

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#12

(Original post by

No because then you would have to know longitude/latitude and the time and date.

**dentistry1**)No because then you would have to know longitude/latitude and the time and date.

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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#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.

_Kar.

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**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.

Thanks

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#16

(Original post by

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 .

**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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#17

(Original post by

Yeah, definitely agreeing that there should be an average value somewhere.

**Kareir**)Yeah, definitely agreeing that there should be an average value somewhere.

**_Kar.**__I refuse to believe you do not either have a formula, or access to a value for this.__
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#18

(Original post by

Ok after i find how much energy hits the surface of the earth from the sun what formula do i use? Please help

Thanks

**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

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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(Original post by

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!

**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

Thanks for your help

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#20

(Original post by

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

Thanks for your help

**mujahid_e3**)Ok if you can just post the answer and i will see your working out and see if i understand

Thanks for your help

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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