Question

Part A: The sun produces 3.9 ⋅ 10^33 ergs of radiant energy per second. How many ergs of radiant energy does the sun produce in 1.55 ⋅ 10^7 seconds? (5 points)

Part B: Which is the more reasonable measurement of the distance between the tracks on a CD:
1.6 ⋅ 10^−3 mm or 1.6 ⋅ 10^3 mm? Justify your answer. (5 points)

Answer 1

@Teddyiswatshecallsme @perl

Answer 2

In order to help guide and walk you through this problem, think of how to get radiants of energy with the sun's production of radiant energy rate. What is the rate formula? How do you find the total output with the rate formula? By not telling you the direct answer and helping give you hints to walk you through this problem, you will achieve your goal of understanding how this problem works.

Answer 3

( 3.9 x 10^33 ) x ( 1.55 x 10^7 ) = ( 3.9 x 1.55 ) x 10^( 33 + 7 ) = 6.045 x 10^40
So 6.045 x 10^40.

? @mhchen

Answer 4

I'd say you did that correctly

Answer 5

sure? lol

Answer 6

Yes
Also why the measure of time 1.55 *10^7?
That's roughly a half year.

Answer 7

ok thanks

Answer 8

can you help me check my answers..theres three

Answer 9

Okay

Answer 10

Answers to question 1 are correct
Going to check question 2

Answer 11

ok

Answer 12

hows the second one?

Answer 13

I get an answer of 4

I)
(3^8 * 2^-5 *9^0)^-2 =
(205)^-2 = 1/(205)^2 =
2.3788^-5

II)
(2^-2/3^3)^4 =
(.25/27)^4 =
(0.0092592593)^4 =
7.3503^-9

III)
3^38 = 2.2877^13

I * II * III =

=2.3788^-5 * 7.3503^-9 * 2.2877^13

= 4

Correct again !!

Answer 14

I need help with part b of the last one though....maybe 1.6 x 10^-3?

Answer 15

Yes 1.6 * 10^-3

If the spacing of the tracks were 1.6*10^3mm that would mean the space between each track would be 1.6 METERS which is much larger than a CD.

Answer 16

ok

Answer 17

Glad I could help

Answer 18

thanks!!:)

Answer 19

wait is 3 right lol i forgot that one:)