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

Answer

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.
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