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Mathematics 17 Online
OpenStudy (anonymous):

An 70 kg weight watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 790 kcal (1 kcal = 1000 cal). How high must the person climb?

OpenStudy (anonymous):

what's the formula that relates the calories to burn and the weight of the person with the steepness of the mountain and how much to climb? is there one?

OpenStudy (anonymous):

i didn't receive one on my worksheet , hope someone knows

OpenStudy (anonymous):

ok thanks I'll leave it for someone else

OpenStudy (anonymous):

you must first convert Cal to cal. 1 Cal=1kcal, then convert cal to joules

OpenStudy (anonymous):

im not sure how to

OpenStudy (colouringme):

1 cal = 4.2 joule

OpenStudy (astrophysics):

W = mgh (potential energy)

OpenStudy (astrophysics):

\[(1000\times10^3~calories) \times~4.2 J\]

OpenStudy (astrophysics):

\[\large Height = \frac{ 4.2\times10^6 }{ 70kg \times9.8\frac{ m }{ s^2 } }=?\]

OpenStudy (anonymous):

this is incorrect

OpenStudy (astrophysics):

You'll get 6.12x10^3 m.

OpenStudy (astrophysics):

How so?

OpenStudy (anonymous):

790 Cal, not 1000 Cal

OpenStudy (astrophysics):

Oh, my bad, just make that 790x10^3 calories, same process.

OpenStudy (anonymous):

please guide him, don't just solve it. make sure he understands each step

OpenStudy (astrophysics):

Pls.

OpenStudy (anonymous):

you can't even read a question right. pls yourself

OpenStudy (astrophysics):

Btw, I love your guide.

OpenStudy (anonymous):

he's right tho, we do need to guide them, i got a warning from the moderators for handing out the answers straight away... some users here dont want to learn at all just want the answer and it can get hard to teach them anything, so what do u do then? give the answer or keep hitting the brick wall :)

OpenStudy (astrophysics):

I did guide him, I showed him the formula, and you can see I didn't give the final answer since I messed up on one number, only reason eHoaX said that was because he couldn't follow the process himself. I'm sure the OP can follow it easily, since I mentioned W=mgh. That's all OP needs, and after that, it's simple.

OpenStudy (anonymous):

im not sure what is the final equation?

OpenStudy (anonymous):

Did you find out the energy in that particular amount of chocolate?

OpenStudy (anonymous):

im not sure how to solve for anything the equation that was shown to my is confusing.

OpenStudy (anonymous):

Okay first you find the amount f energy in that chocolate 790 *10 ^3 * 4.2 that would give you the amount of energy in that chocolate Good now?

OpenStudy (anonymous):

is the answer 11060. (does "^" mean divide?)

OpenStudy (anonymous):

to the power = ^

OpenStudy (anonymous):

or \[790\times1000\times4.2\]

OpenStudy (anonymous):

3318000?

OpenStudy (anonymous):

yes good Now what is the formula for the increase in potential energy when climbing a distance h

OpenStudy (anonymous):

w=mgh ?

OpenStudy (anonymous):

correct now you know W , g and m just plug in the values and find h :)

OpenStudy (anonymous):

W-3318000 G-70 M-790 H- Solve for ?

OpenStudy (anonymous):

g = 9.8 or 10 M = 70 W = 3318000 solve for H

OpenStudy (anonymous):

464520 ?

OpenStudy (anonymous):

how did u get it

OpenStudy (anonymous):

3318000/70*9.8

OpenStudy (anonymous):

then you should be getting 4836

OpenStudy (anonymous):

im still getting 464520....am i multiplying something wrong ?

OpenStudy (anonymous):

\[h = \frac{3318000}{70\times9.8}\] make sure u bracket ur below values

OpenStudy (anonymous):

\[h=\frac{3318000}{(70\times9.8)}\]

OpenStudy (anonymous):

4836.7346?

OpenStudy (anonymous):

yes :)

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