anonymous
  • anonymous
A astronaut has 100 kg of mass. Unfortunately the mass of his body was spread out in a sphere. After that, the density inside the sphere is 2x10^-28 kg/m^3. What is the radius of the sphere where the mass of the astronaut was spread out?
Physics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Density = MASS/VOLUME
anonymous
  • anonymous
use it to find the volume of the sphere
anonymous
  • anonymous
I just set up the following relation 6x10^-28=100kg/V

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yes u r correct. PLZ CONTINUE
anonymous
  • anonymous
So, with some algebric manipulation i found this relation V=1x10^2/6x10^-28 and i got V=1/6x10^30.
anonymous
  • anonymous
I think this is the volume of the sphere.
anonymous
  • anonymous
Now, Volume of a sphere=4/3 pi r^3
anonymous
  • anonymous
Use it to find r
anonymous
  • anonymous
I just plug it in the formual, right?
anonymous
  • anonymous
YEp, 1/6x10^30 = 4/3 pi r^3
anonymous
  • anonymous
got it?
anonymous
  • anonymous
Just one moment.
anonymous
  • anonymous
i got 1,46459
anonymous
  • anonymous
But I got different
anonymous
  • anonymous
What do you got?
anonymous
  • anonymous
i write the equation in this order. cubic root of 1/6x10^30=4/3phi*r
anonymous
  • anonymous
Now i got 4,12740906x10^9 m
anonymous
  • anonymous
Density= MASS/ VOLUME 2x10^-28=100/VOLUME VOLUME=50 * 10 ^28 4/3 pi r^3 = 50 * 10^28 r^3=4923725109
anonymous
  • anonymous
That ´s right.
anonymous
  • anonymous
r=4923725109
anonymous
  • anonymous
How do u calculate the 2x10^-28=100/VOLUME. Plz do it step by step.
anonymous
  • anonymous
DEnsity =MASS/ VOLUME
anonymous
  • anonymous
Now, substitute the given values
anonymous
  • anonymous
thanks
anonymous
  • anonymous
:)yw
anonymous
  • anonymous
But sorry i can´t understant how do you calculate 2x10^-28=100/VOLUME. May you do it plz?
anonymous
  • anonymous
When i calculate V i have found 50/3x10^28. How did you got 50 * 10 ^28. PLz.
anonymous
  • anonymous
100/2 =50 right?
anonymous
  • anonymous
Yes, but where did you find the number two? If we have the division 100/6?
anonymous
  • anonymous
Did you divide 6/3. I got it. now.

Looking for something else?

Not the answer you are looking for? Search for more explanations.