Part A: The sun produces 3.9 ⋅ 1033 ergs of radiant energy per second. How many ergs of radiant energy does the sun produce in 3.25 ⋅ 103 seconds? (5 points)
Part B: Which is the more reasonable measurement of the distance between the tracks on a railroad:
1.435 ⋅ 10−3 mm or 1.435 ⋅ 103 mm? Justify your answer

- anonymous

- schrodinger

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- anonymous

help please

- anonymous

Should that say
The sun produces \( 3.9 ⋅ 10^{33} \) ergs of radiant energy per second

- anonymous

yes @jayzdd

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## More answers

- anonymous

$$\large \sf \frac{3.9 ⋅ 10^{33} ~ergs} { second} \times 3.25 ⋅ 10^3 ~ seconds$$

- anonymous

ok is there anything else to that?

- anonymous

@jayzdd ^

- anonymous

now we get out our calculators :)

- anonymous

alright so i calculate that?

- anonymous

yes

- anonymous

alright hold on

- anonymous

i dont know if this is right but i got 1.2675 × 10^30

- anonymous

close

- anonymous

$$ 1.2675 × 10^{37}$$

- anonymous

ohhh ok is that the final answer?

- anonymous

yes and you can round it

- anonymous

ok thank you, can you help with part B also please?

- anonymous

which measurement seems more likely?

- anonymous

um 1.435 ⋅ 10^3 mm?

- anonymous

yes

- anonymous

1 mm is very small , so 10^-3 mm is extremely tiny

- anonymous

right thank you! i have one more question that i have no idea how to work on, can you help me?

- anonymous

okay

- anonymous

Find the value of the following expression:
(2^8 ⋅ 3^−5 ⋅ 6^0)^−2 ⋅ 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 2^28 (5 points)
Write your answer in simplified form. Show all of your steps

- anonymous

????

- anonymous

Let's rewrite this long expression

- anonymous

ok

- anonymous

how do i re write it

- anonymous

alright so can you help me solve it

- anonymous

\[ (2^8 ⋅ 3^{-5} ⋅ 6^0)^{−2} ⋅ \frac{3^{-2}}{2^3} 4 ⋅ 2^28\]

- anonymous

alright so the one is parentheses are first correct?

- anonymous

I am not sure about the expression 'whole'

- anonymous

the 4 by the fraction is in square root

- anonymous

got it?

- Spring98

Is there any answer to the last one?

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