One cubic meter (1.00 m3) of aluminum has a mass of 2.70 × 103 kg, and 1.00 m3 of iron has a mass of 7.86 × 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.00 cm on an equal-arm balance
Added by Elisa T.
Step 1
The formula for the volume of a sphere is V = 4/3πr³. Substituting the given radius of 2.00 cm (or 0.02 m) into the formula, we get: V_iron = 4/3π(0.02 m)³ = 3.35 x 10^-5 m³ Show more…
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One cubic meter (1.00 m3) of aluminum has mass of 2.70 x 103 kg, and the same volume of iron has mass of 7.86 x 103 kg Find the radius of solid aluminum sphere that will balance solid iron sphere of radius 4.18 cm on an equal-arm balance.
Shaiju T.
One cubic meter $\left(1.00 \mathrm{m}^{3}\right)$ of aluminum has a mass of $2.70 \times 10^{3} \mathrm{kg},$ and the same volume of iron has a mass of $7.86 \times 10^{3} \mathrm{kg} .$ Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius $2.00 \mathrm{cm}$ on an equal-arm balance.
Dis One cubic meter $\left(1.00 \mathrm{m}^{3}\right)$ of aluminum has a mass of $2.70 \times 10^{3} \mathrm{kg}$ , and the same volume of iron has a mass of $7.86 \times 10^{3} \mathrm{kg}$ . Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.00 $\mathrm{cm}$ on an equal-arm balance.
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