Solving an equation which should give an animal's foraging radius

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The discussion centers around solving an equation to determine an animal's foraging radius based on energy and velocity parameters. The original formula presented appears to have a typographical error, leading to incorrect calculations by participants. A key correction suggests that the equation should include square brackets to clarify the order of operations. Ultimately, it is determined that the value for Ema should be adjusted from 24 Watts to 42 Watts to arrive at the correct answer of 116 km. The conversation highlights the importance of accurate data representation in scientific equations.
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Hi all, I have the following problem which should show me the distance an animal can travel

Homework Statement



e = 7.8*10^6 Joules
V= 12m/s
k= 2.0 Joules/m
Ema = 24 Watts
Emc = 42 Watts

The answer should be 116km

Homework Equations



r = eV/12(Ema + 0.5Emc) + 2kV

The Attempt at a Solution




I'm coming out with 159183.6735 which is certainly wrong.

Thanks
 
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No, I think simply distance = energy / (energy/m) then you will get 3.9 10^6 m. The key is wrong I think...
 
This is given in an academic journal so I'm not sure where I've gone wrong, I've taken the values as they are presented.
 
killbot2000 said:
Hi all, I have the following problem which should show me the distance an animal can travel

Homework Statement



e = 7.8*10^6 Joules
V= 12m/s
k= 2.0 Joules/m
Ema = 24 Watts
Emc = 42 Watts

The answer should be 116km

Homework Equations



r = eV/12(Ema + 0.5Emc) + 2kV

The Attempt at a Solution

I'm coming out with 159183.6735 which is certainly wrong.

Thanks

Firstly you have certainly written the formula wrong and it is most likely

r = eV/[12(Ema + 0.5Emc) + 2kV]

(which tells you something about how to do the arithmetic operations).

This makes sense in terms of dimensions which without the square brackets it doesn't. (There are other possibilities so use the formula in the journal, not mine if different!)
 
Hi, the way I've written is the way it's given in the journal. Perhaps a typo on their part?
 
killbot2000 said:
Hi, the way I've written is the way it's given in the journal. Perhaps a typo on their part?

It does happen. Perhaps you could reproduce it directly from the journal here? Or link if this is free journal access.

But anyway your kV is (joules/m).(m/s) = joules/s = watts , the same unit as the other things I have grouped it with. It can't stand there on its own being equated to something expressed in meters on the other side of the equation.
 
Unfortunately I can't link directly because it's a hard copy. Perhaps I can give you another example where I seem to have worked out the answer correctly.

r = [e - T(Ema+Emc/2)]/2k

e = 7.8*10^6 Joules
k= 2.0 Joules/m
Ema = 24 Watts
Emc = 42 Watts
T = 48 hours

[ 7.8*10^6 - 172800 (24 + 42/2)]/4

The answer should be 6km and I got from the above
= 6000 m
 
The answers seem to be available.
What did my proposed version give?
Other possible likely version it could just have been

r = eV/12[(Ema + 0.5Emc) + 2kV]
 
r = eV/12(Ema + 0.5Emc) + 2kV

Hi,

My approach was as follows 7.8 x 10^6 x 12/[12(24+42) + 2 x 2 x 12]

That gives 111428.571429
 
  • #10
Hi all,

I think I solved the problem for those interested. It seems to have been a typo on the part of the author. Of the original values Ema should read 42 not 24.
 
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