Question Part Points Submissions Used 1 0/1 3/10 Total 0/1 . .. Using 2.60 1017

Question

Question Part Points Submissions Used 1 0/1 3/10 Total 0/1 . .. Using 2.60 1017 kg/m3 as the density of nuclear matter, find the radius of a sphere of such matter that would have a mass equal to that of Earth. Earth has a mass equal to 5.98 1024 kg and average radius of 6.37 106 m Question Part Points Submissions Used 1 0/1 3/10 Total 0/1 . .. Using 2.60 1017 kg/m3 as the density of nuclear matter, find the radius of a sphere of such matter that would have a mass equal to that of Earth. Earth has a mass equal to 5.98 1024 kg and average radius of 6.37 106 m.

Explanation / Answer

density, rho = 2.6*10^17 kg/m^3

Mass, M = 5.98*10^24 kg

let R would be the radius of earth with the given density.

now Apply, density = mass/volume

Volume = mass/density

(4/3)*pi*R^3 = M/rho

R^3 = (3*M/(4*pi*rho))

R = (3*M/(4*pi*rho))^(1/3)

= (3*5.98*10^24/(4*pi*2.6*10^17))^(1/3)

= 176.4 m <<<<<<=----------Answer

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