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notka56 [123]
3 years ago
9

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 58.2 cm, and the density of iro

n is 7.87 g/cm3. Find the inner diameter.
Physics
1 answer:
vredina [299]3 years ago
3 0

Answer:

0.556m

Explanation:

unit conversion

58.2 cm = 0.582 m -> the outer radius is 0.582/2 = 0.291 m

7.87 g/cm^33 = 7.87 g/cm^33 * 1/1000 (kg/g) * 100^3 cm^3/m^3 = 7870 kg/m^3

Let r be the inner diameter we can find the volume of the hollow part of the sphere using the following formula

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

The volume of the iron shell is the volume of the whole sphere subtracted by volume of the hollow part

V_s = V - V_h = \frac{4}{3}\pi 0.291^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.0246 - r^3)

Let water density \rho_w = 1000 kg/m^3, then the buoyant force is the weight of water displaced by the shell

F_b = g\rho_w V = g1000\frac{4}{3}\pi(0.291^3) = g1000\frac{4}{3}\pi(0.0246)

For the shell to almost completely submerged in water, its buoyant force equal to the weight of the shell

F_b = W_s

g1000\frac{4}{3}\pi(0.0246) = g\rho_sV_s

g1000\frac{4}{3}\pi(0.0246) = g7870\frac{4}{3}\pi(0.0246 - r^3)

1000*0.0246 = 7870(0.0246 - r^3)

24.6 = 193.602 - 7870r^3

r^3 = (193.602 - 24.6)/7870 = 0.02147

r = \sqrt[3]{0.02147} = 0.278 m

So the inner diameter is 0.278*2 = 0.556 m

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1)

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