Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 1 - Introduction, Measurement, Estimating - General Problems: 61


There is $3.68\times10^{51}~kg$ of "ordinary" matter in the observable universe.

Work Step by Step

First let's calculate the volume of the observable universe: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (13.0\times10^{25})^3 = 9.20\times10^{78}~m^3$ The density $\rho = 1\times10^{-26}~kg/m^3$ $mass = V\times\rho = (9.20\times10^{78}~m^3)(1\times10^{-26}~kg/m^3) = 9.20\times10^{52}~kg$ The "ordinary" mass is only 4% of this total mass, so the "ordinary" mass is: $0.04\times (9.20\times10^{52}~kg) = 3.68\times10^{51} ~kg$ There is $3.68\times10^{51}~kg$ of ordinary matter in the observable universe.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.