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one cubic meter(1.00m³) of aluminum has a mass of 2.70 x 10³kg and 1.00 m³ of iron has a mass of 7.86 x 10³ kg. find the radius of solid aluminum sphere that balance a solid iron sphere of radius 2.00cm on an equal arm balance​

Sagot :

Answer:

To find the radius of the solid aluminum sphere that balances a solid iron sphere of radius 2.00 cm on an equal-arm balance, we need to calculate the mass of the iron sphere and then determine the radius of the aluminum sphere that has the same mass.

Given:

- Mass of iron sphere = 7.86 x 10³ kg

- Radius of iron sphere = 2.00 cm = 0.02 m

First, let's calculate the mass of the iron sphere using the formula for the volume of a sphere:

Volume of a sphere = (4/3) * π * r³

Where:

- r is the radius of the sphere

Mass of iron sphere = Density of iron * Volume of iron sphere

The density of iron is approximately 7.86 x 10³ kg/m³. Plugging in the values:

Mass of iron sphere = (7.86 x 10³ kg/m³) * ((4/3) * π * (0.02 m)³)

Mass of iron sphere = 7.86 x 10³ kg

Now, we need to find the radius of the aluminum sphere that has the same mass. The density of aluminum is approximately 2.70 x 10³ kg/m³. We can use the same formula to calculate the volume of the aluminum sphere:

Volume of aluminum sphere = (4/3) * π * r³

Mass of aluminum sphere = Density of aluminum * Volume of aluminum sphere

Now, we can set up an equation to find the radius of the aluminum sphere:

(7.86 x 10³ kg) = (2.70 x 10³ kg/m³) * ((4/3) * π * r³)

Solving for r³:

r³ = (7.86 x 10³ kg) / (2.70 x 10³ kg/m³)

r³ = 2.91 x 10⁰ m³

Taking the cube root of both sides to find the radius of the aluminum sphere:

r = (2.91 x 10⁰ m³)^(1/3)

r ≈ 1.70 m

Therefore, the radius of the solid aluminum sphere that balances a solid iron sphere of radius 2.00 cm on an equal-arm balance is approximately 1.70 m.

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