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lesya [120]
3 years ago
15

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

bers plugged in with proper UNITS, and clearly indicate your final answer).
Mass=1.3x1022kg, Diameter=2300km

B) (extra credit problem) Working backwards... Calculate the mean radius of the following object (if it was a perfect sphere).

Density=687kg/m3, Mass=5.68x1026kg
Chemistry
1 answer:
-BARSIC- [3]3 years ago
3 0

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}

V = \frac{ 4}{3 }\pi r^{3}

r^{3} = \frac{3V }{4 \pi }\

r= \sqrt [3]{ \frac{3V }{4 \pi } }

r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}

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