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# physics

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The mass of tithe earth is 5.98x10^24kg, and its radius is  6.38x10^8m. Compute the density of the earth, using powers-of-ten notation and the correct number of significant figures.

Sep 14, 2021

#1
+13017
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The mass of tithe earth is 5.98x10^24kg, and its radius is 6.38x10^8m. Compute the density of the earth, using powers-of-ten notation and the correct number of significant figures.

Hello Guest!

The radius of our earth is $$6393km \approx 6.39 \times10 ^ 6m$$.

$$Density=\dfrac{mass}{volume}$$

$$\rho=\dfrac{m}{V}=\dfrac{5.98\cdot 10^{24}\cdot kg}{\frac{4}{3}\pi\cdot 6.39^3\cdot (10^6)^3\cdot m^3}\times\dfrac{m^3}{10^6\cdot cm^3}\times\dfrac{10^3\cdot g}{kg}\\ =\dfrac{3\cdot 5.98}{4\cdot 6.39^3\cdot \pi}\times\dfrac{10^{24}\cdot 10^3}{10^{18}\cdot 10^6}\times\dfrac{g}{cm^3}\color{blue}\approx5.47g/cm^3$$

!

Sep 14, 2021
#2
+35290
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Using different units:

m / v = density

5.98* 10^24 kg  /  4/3 pi (6.38* 10^8)^3   = .0054973  kg/m3   =5.49 *10-3  kg/m3

Sep 14, 2021
edited by ElectricPavlov  Sep 14, 2021
#3
+13017
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Where is this "tithe earth" whose density is $$5.5g / m^3$$, what do you want to say?

asinus  Sep 14, 2021