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Maksim231197 [3]
3 years ago
12

Assuming that 70 percent of the Earth’s surface is covered with water at average depth of 0.95 mi, estimate the mass of the wate

r on Earth. One mile is approximately 1.609 km and the radius of the earth is 6.37 × 10^6 m. Answer in units of kg. Your answer must be within ± 20.0%
Physics
1 answer:
Alla [95]3 years ago
8 0

Answer:

The mass of the water on earth is 5.4537\times10^{20}\ kg

Explanation:

Given that,

Average depth h= 0.95 mi

h=0.95\times1.609

h =1.528\ km

h=1.528\times10^{3}\ m

Radius of earth r= 6.37\times10^{6}\ m

Density = 1000 kg/m³

We need to calculate the area of surface

Using formula of area

A =4\pi r^2

Put the value into the formula

A=4\pi\times(6.37\times10^{6})^2

A=5.099\times10^{14}\ m^2

We need to calculate the volume of earth

V = Area\times height

Put the value into the formula

V=5.099\times10^{14}\times1.528\times10^{3}

V=7.791\times10^{17}\ m^3

Now, 70 % volume of the total volume

V= 7.791\times10^{17}\times\dfrac{70}{100}

V=5.4537\times10^{17}\ m^3

We need to calculate the mass of the water on earth

Using formula of density

\rho = \dfrac{m}{V}

m = \rho\times V

Put the value into the formula

m=1000\times5.4537\times10^{17}

m =5.4537\times10^{20}\ kg

Hence, The mass of the water on earth is 5.4537\times10^{20}\ kg

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