Physics word problem(multiplying and dividing powers)

[Walker]

the class is a beginners astronomy course and the math is supposed to be simple and done to the nearest powers of ten but I've never multiplied or divided numbers like these in my life. any advice? i dont need to be an expert, there isnt any math on the exams, this is just a big one off assignment.

I put the number representing the powers in () form, and wrote pie to represent the pie symbol which i cant find on my keyboard.

Approximate dust grains as spheres, each with a radius of R = 10(−5) cm. Each grain of dust presents a cross-sectional shadow of pieR(2).

Recall that a photon’s mean-free-path, L, is the typical distance a photon flies before bumping into matter.

If the number density of dust grains (number of grains per cm(3) ) is n, the mean-free-path length relationship to the cross-sectional shadow is:

L = 1 divided by npieR2

Given that interstellar extinction observations indicate that L = 3 × 10(3) lightyears, calculate the number density, n, of dust grains in units of cm(−3) . How many dust grains would you expect to find in a volume of a cube about 100 metres in all directions?

my work so far (im a history major and completely out of my element here...)


3 X 10(3) = 1 divided by n pie 10(-5)(2) how do i solve for n when 10(-5)(2) has 2 powers attached to it, are the powers multiplied by each other? ie -5*2 = 10(-10) ?

(the problem is question 1A on this assignment, if seeing the equations properly helps) http://www.maybury.ca/phys1902/documents/assignment2.pdf

 

[HallsofIvy]

the class is a beginners astronomy course and the math is supposed to be simple and done to the nearest powers of ten but I've never multiplied or divided numbers like these in my life. any advice? i dont need to be an expert, there isnt any math on the exams, this is just a big one off assignment.

I put the number representing the powers in () form, and wrote pie to represent the pie symbol which i cant find on my keyboard.

More common is "^" to indicate powers. Your "10(5)" would be written "10^5". Also the standard transliteration of the Greek letter "\(\displaystyle \pi\)" is "pi", not "pie".

Approximate dust grains as spheres, each with a radius of R = 10(−5) cm. Each grain of dust presents a cross-sectional shadow of pieR(2).

R= 10^(-5) and the "cross-sectional shadow" is \(\displaystyle \pi (10^{-5})^2= \pi (10)^{-10}= 0.0000000003141592...\).

Recall that a photon’s mean-free-path, L, is the typical distance a photon flies before bumping into matter.

If the number density of dust grains (number of grains per cm(3) ) is n, the mean-free-path length relationship to the cross-sectional shadow is:

L = 1 divided by npieR2

\(\displaystyle \frac{1}{n\pi R^2}=\frac{3183098862}{n}\).

Given that interstellar extinction observations indicate that L = 3 × 10(3) lightyears, calculate the number density, n, of dust grains in units of cm(−3) . How many dust grains would you expect to find in a volume of a cube about 100 metres in all directions?

So 3 x 10^3= 3000= 318098862/n. Solve for n. Since a meter is 100 cm, "a cube about 100 m in all directions" is "a cube about 10000 cm in all directions" so has a volume of (10000)^3= 1000000000000 cubic cm.

my work so far (im a history major and completely out of my element here...)


3 X 10(3) = 1 divided by n pie 10(-5)(2) how do i solve for n when 10(-5)(2) has 2 powers attached to it, are the powers multiplied by each other? ie -5*2 = 10(-10) ?

Yes, \(\displaystyle (10^{-5})^2= 10^{-10}\).

(the problem is question 1A on this assignment, if seeing the equations properly helps)

http://www.maybury.ca/phys1902/documents/assignment2.pdf

 

[Walker]

checking my final answer

More common is "^" to indicate powers. Your "10(5)" would be written "10^5". Also the standard transliteration of the Greek letter "\(\displaystyle \pi\)" is "pi", not "pie".


R= 10^(-5) and the "cross-sectional shadow" is \(\displaystyle \pi (10^{-5})^2= \pi (10)^{-10}= 0.0000000003141592...\).


\(\displaystyle \frac{1}{n\pi R^2}=\frac{3183098862}{n}\).


So 3 x 10^3= 3000= 318098862/n. Solve for n. Since a meter is 100 cm, "a cube about 100 m in all directions" is "a cube about 10000 cm in all directions" so has a volume of (10000)^3= 1000000000000 cubic cm.


Yes, \(\displaystyle (10^{-5})^2= 10^{-10}\).

thanks for clearing up the powers into powers problem I had, I'm still a little unsure of how to solve for n in this case, the prof said I shouldn't need a calculator for these equations so I was wondering if this would be approximately correct or am i getting an integer off?

318098862/n = 3 x 10^-9 = n (i counted 8 numbers after the 3)

also, i was supposed to calculate the number of dust grains per cm^3 (n) into cm^-3 and I'm unsure what this means.. do the cm calculations cancel each other out? is it the size of an individual dust grain? if so then would my calculation for a 100m cube be the following...

n (in cm^-3) = 3 x 10^-9^-3 = 3 x 10^27? isn't this number too big? and does it represent the amount of dust in a single cubed centimetre?

if it does can I assume then multiplying it ten thousand times(like you mentioned is in the cube) gives me the final answer of dust in a 100-meter cube? ie...

3 x 10^27 / 1 x 10^4 = 3 x 10^23 grains of dust in a 100 meter cube. oooor... did I do it backwards lol I don't know...
http://www.maybury.ca/phys1902/documents/assignment2.pdf