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Chapter 1: Problem 71

A star is estimated to have a mass of \(2 \times 10^{36} \mathrm{kg} .\) Assuming it to be a sphere of average radius \(7.0 \times 10^{5} \mathrm{km}\) , calculate the average density of the star in units of grams per cubic centimeter.

Short Answer

Expert verified
The average density of the star is approximately \(1.39 \times 10^3\) grams per cubic centimeter.

Step by step solution

01

Convert radius to meters

Given radius = \(7.0 \times 10^5 \mathrm{km}\). To convert this to meters, multiply it by 1000: Radius = \(7.0 \times 10^5 \times 1000 \mathrm{m} = 7.0 \times 10^8 \mathrm{m}\).
02

Calculate the volume of the star

Use the formula for the volume of a sphere, \[volume = \frac{4}{3}\pi r^{3}\] Plug in the values: \[volume = \frac{4}{3} \pi (7.0 \times 10^8)^{3} = 1.44 \times 10^{27} \mathrm{m^3} \]
03

Calculate the density of the star

Now, use the density formula, \[density = \frac{mass}{volume}\] Plug in the values: \[density = \frac{2 \times 10^{36} \mathrm{kg}}{1.44 \times 10^{27} \mathrm{m^3}} = 1.39 \times 10^{9} \frac{\mathrm{kg}}{\mathrm{m^3}}\]
04

Convert the density to grams per cubic centimeter

To convert the density to grams per cubic centimeter, first convert kilograms to grams (1 kg = 1000 g) and cubic meters to cubic centimeters (1 m³ = \(10^6\) cm³): \[density = 1.39 \times 10^9 \frac{\mathrm{kg}}{\mathrm{m^3}} \times \frac{1000 \frac{\mathrm{g}}{\mathrm{kg}}}{10^6 \frac{\mathrm{cm^3}}{\mathrm{m^3}}} = 1.39 \times 10^{3} \frac{\mathrm{g}}{\mathrm{cm^3}}\] The average density of the star is approximately \(1.39 \times 10^3\) grams per cubic centimeter.

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Most popular questions from this chapter

The radius of a neon atom is \(69 \mathrm{pm},\) and its mass is \(3.35 \times\) \(10^{-23} \mathrm{g} .\) What is the density of the atom in grams per cubic centimeter \(\left(\mathrm{g} / \mathrm{cm}^{3}\right) ?\) Assume the nucleus is a sphere with volume \(=\frac{4}{3} \pi r^{3}\)

Perform the following mathematical operations, and express the result to the correct number of significant figures. a. \(6.022 \times 10^{23} \times 1.05 \times 10^{2}\) b. $\frac{6.6262 \times 10^{-34} \times 2.998 \times 10^{8}}{2.54 \times 10^{-9}}$ c. \(1.285 \times 10^{-2}+1.24 \times 10^{-3}+1.879 \times 10^{-1}\) d. \(\frac{(1.00866-1.00728)}{6.02205 \times 10^{23}}\) e. $\frac{9.875 \times 10^{2}-9.795 \times 10^{2}}{9.875 \times 10^{2}} \times 100(100 \text { is exact) }$ f. $\frac{9.42 \times 10^{2}+8.234 \times 10^{2}+1.625 \times 10^{3}}{3}(3 \text { is exact) }$

The scientific method is a dynamic process. What does this mean?

Perform the following unit conversions. a. Congratulations! You and your spouse are the proud parents of a new baby, born while you are studying in a country that uses the metric system. The nurse has informed you that the baby weighs 3.91 kg and measures 51.4 cm. Convert your baby’s weight to pounds and ounces and her length to inches (rounded to the nearest quarter inch). b. The circumference of the earth is 25,000 mi at the equator. What is the circumference in kilometers? in meters? c. A rectangular solid measures 1.0 m by 5.6 cm by 2.1 dm. Express its volume in cubic meters, liters, cubic inches, and cubic feet.

How many significant figures are in each of the following? a. 100 b. \(1.0 \times 10^{2}\) c. \(1.00 \times 10^{3}\) d. 100 e. 0.0048 f. 0.00480 g. \(4.80 \times 10^{-3}\) h. \(4.800 \times 10^{-3}\)
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