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If formula E=(1.74x10^19 x 10^1.44M)

I am trying to do this question based on recall memory alone. I sense I’m forgetting an important formula regarding initial energy, or am I confusing that with mechanics in terms of like a cars initial velocity..

Do ignore E here or is that part of the equation.

I left this to last because I've been trying to do it by myself but I am a little confused

See attached:

IMG_0999.jpeg

Or, do I leave in the form E=1.74(10²⁶)
(edited 1 month ago)
Original post by KingRich
I am trying to do this question based on recall memory alone. I sense I’m forgetting an important formula regarding initial energy, or am I confusing that with mechanics in terms of like a cars initial velocity..
Do ignore E here or is that part of the equation.
I left this to last because I've been trying to do it by myself but I am a little confused
See attached:
IMG_0999.jpeg
Or, do I leave in the form E=1.74(10²⁶)

The only formula needed is the one provided.
For ii)
Set E to 2(1.74 * 10^26.2) and rearrange for M.
Log(a x b) = log(a) + log(b)

If you wished to use your logE answer by doubling, before you substitute,
You would have to log:
E = 1.74 * 10^19 * 10^1.44M

Are you sure you logged correctly?
Log(a * b) = log(a) + log(b)IMG_5636.jpeg
(edited 1 month ago)
Reply 2
Original post by BankaiGintoki
The only formula needed is the one provided.
For ii)
Set E to 2(1.74 * 10^26.2) and rearrange for M.
Log(a x b) = log(a) + log(b)
If you wished to use your logE answer by doubling, before you substitute,
You would have to log:
E = 1.74 * 10^19 * 10^1.44M
Are you sure you logged correctly?
Log(a * b) = log(a) + log(b)IMG_5636.jpeg

Oh, do you mean by ignoring the little .2? It was just for keepsake of keeping it a little cleaner…

So as part i) leaving as 1.74 x 10^26.2 is absolutely fine. Am I not meant to apply log to find a true value? Such as the calculator says, 2.757714155x₁₀²⁶. Or is kinda impossible?

I’ve answered part 2 and again. However; the -19 looks a little out of place to me. But I know that you can’t have a negative log. Anyway: see below

IMG_1003.jpeg


I should add that this is for an access course, so I don't know how advance they would expect me to be.
(edited 1 month ago)
Original post by KingRich
Oh, do you mean by ignoring the little .2? It was just for keepsake of keeping it a little cleaner…
So as part i) leaving as 1.74 x 10^26.2 is absolutely fine. Am I not meant to apply log to find a true value? Such as the calculator says, 2.757714155x₁₀²⁶. Or is kinda impossible?
I’ve answered part 2 and again. However; the -19 looks a little out of place to me. But I know that you can’t have a negative log. Anyway: see below
IMG_1003.jpeg
I should add that this is for an access course, so I don't know how advance they would expect me to be.

IMG_5637.jpegIMG_5639.jpeg
IMG_5640.jpeg

Yours is correct.IMG_5642.jpeg
(edited 1 month ago)
Reply 4
Original post by BankaiGintoki
IMG_5637.jpegIMG_5639.jpeg
IMG_5640.jpeg
Yours is correct.IMG_5642.jpeg

Fantastic. I shall leave as it is then because without introducing a calculator, I can’t possibly find a true value.

Thank you!!
Reply 5
Original post by KingRich
Fantastic. I shall leave as it is then because without introducing a calculator, I can’t possibly find a true value.
Thank you!!

Cant help think youre overcomplicating it somewhat (rabbit in headlights with large numbers) though to get a number at the end you do need a simple log calculation.

For a) it would make sense just to write it in scientific notation so
(1.74*10^0.2)*10^26
where you use your calculator to evaluate the mantissa (1.74*10^0.2) and lets just say the answer is
m * 10^26
where m lies between 1 and 10 (the usual assumption).

For b), the new richter number just be a bit greater than 5 to double the energy so lets say its 5+N, then
m*10^(26+1.44N) = 2m * 10^26
cancel common terms and you should get
10^1.44N = ...
take logs base 10
N = ...
Then obv the value is 5+N

A duffers way to get N would be to note
2^3 ~ 10
so
10^0.3 ~ 2
so
10^(1.44*0.2) ~ 2
So N ~ 0.2 and the new richter value would be ~5.2
(edited 1 month ago)
Reply 6
Original post by mqb2766
Cant help think youre overcomplicating it somewhat (rabbit in headlights with large numbers) though to get a number at the end you do need a simple log calculation.
For a) it would make sense just to write it in scientific notation so
(1.74*10^0.2)*10^26
where you use your calculator to evaluate the mantissa (1.74*10^0.2) and lets just say the answer is
m * 10^26
where m lies between 1 and 10 (the usual assumption).
For b), the new richter number just be a bit greater than 5 to double the energy so lets say its 5+N, then
m*10^(26+1.44N) = 2m * 10^26
cancel common terms and you should get
10^1.44N = ...
take logs base 10
N = ...
Then obv the value is 5+N

Yeah, it does seem the case that I do overthink in some questions. For I) you’ve extracted the 0.2 from the power. In that just for an example? Also, what’s a mantissa?

Ah, it’s been a while. So, as we now know E when m=5....

5+n = where n to be found is the additional value where E is doubled.

I'm not sure how you find m*10*(26+1.44N)

I shall see what my tutor says after my initial draft as it's allowed on this occasion.

I'll revisit this if I need to. In fact I'll have another look when I get home and see if I can figure it out lol
Reply 7
Original post by KingRich
Yeah, it does seem the case that I do overthink in some questions. For I) you’ve extracted the 0.2 from the power. In that just for an example? Also, what’s a mantissa?
Ah, it’s been a while. So, as we now know E when m=5....
5+n = where n to be found is the additional value where E is doubled.
I'm not sure how you find m*10*(26+1.44N)
I shall see what my tutor says after my initial draft as it's allowed on this occasion.
I'll revisit this if I need to. In fact I'll have another look when I get home and see if I can figure it out lol
You can google scientific notation and its
m * 10^n
say, where n is an integer and |m| lies between 1 and 10. You have 10^26.2 = 10^0.2 * 10^26 so m is as above.

For b) you want to double the previous calculation. This will happen if 5 is increased a bit to 5+N say then you have
previous energy * 10^(1.44N) = 2 * previous energy
and previous energy cancels and you can easily solve for N (simple log relationship). Its well known that 10^0.3 ~ 2 (or 2^3 ~ 10) so really you want N to be about 0.2.

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