One cubic meter (1.00 m^3) of aluminum has a mass of 2.70 multiplied by 10^3 kg, and the same volume of iron has a mass of 7.86 multiplied by 10^3 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.40 cm on an equal-arm balance.
To have the same weight and mass, the product of Radius^3 and density must be the same for the aluminum and iron spheres.
Let subscript 1 be Al and subscript 2 be iron.
R1^3*(density1) = R2^3*(density2)
(R1/R2)^3 = 7.86/2.70 = 2.911
R1/R2 = cube root of 2.911 = 1.428
R1 = 1.40*1.428 = 2.00 meters
(using three significnant figures)
To find the radius of the solid aluminum sphere, we need to use the concept of density.
Density is defined as mass divided by volume:
Density = mass / volume
For aluminum:
Density of aluminum = mass of aluminum / volume of aluminum
For iron:
Density of iron = mass of iron / volume of iron
Since the two spheres are balanced on an equal-arm balance, their densities must be equal:
Density of aluminum = Density of iron
Using the previously mentioned density formula, we have:
mass of aluminum / volume of aluminum = mass of iron / volume of iron
Let's plug in the given values:
(2.70 x 10^3 kg) / volume of aluminum = (7.86 x 10^3 kg) / volume of iron
Since the volume of a sphere is given by (4/3)πr^3, we can write:
(2.70 x 10^3 kg) / [(4/3)πr^3] = (7.86 x 10^3 kg) / [(4/3)π(0.014 m)^3]
Simplifying the equation:
(2.70 x 10^3) / (4/3)πr^3 = (7.86 x 10^3) / (4/3)π(0.014^3)
Canceling out common factors:
2.70 / r^3 = 7.86 / (0.014^3)
Cross-multiplying:
(2.70)(0.014^3) = (7.86)(r^3)
Rearranging the equation:
r^3 = (2.70)(0.014^3) / 7.86
Calculating the expression on the right-hand side:
r^3 ≈ (0.056588)(0.000274) / 7.86
r^3 ≈ 0.00000001548 / 7.86
r^3 ≈ 1.968 x 10^-12
Taking the cube root of both sides:
r ≈ ∛(1.968 x 10^-12)
r ≈ 0.000128 m
Therefore, the radius of the solid aluminum sphere that will balance a solid iron sphere of radius 1.40 cm on an equal-arm balance is approximately 0.000128 meters or 0.128 mm.
To find the radius of a solid aluminum sphere that will balance a solid iron sphere on an equal-arm balance, we need to use the principle of balance. The principle states that the torque produced by the aluminum sphere is equal to the torque produced by the iron sphere.
The torque produced by a sphere is given by the formula: torque = force × perpendicular distance.
Let's start by finding the force produced by each sphere. The force is the weight of the sphere, which is given by the formula: force = mass × acceleration due to gravity.
For the aluminum sphere:
Mass = 2.70 × 10^3 kg
Acceleration due to gravity = 9.8 m/s^2
Force(aluminum) = Mass × Acceleration due to gravity = (2.70 × 10^3 kg) × (9.8 m/s^2)
For the iron sphere:
Mass = 7.86 × 10^3 kg
Acceleration due to gravity = 9.8 m/s^2
Force(iron) = Mass × Acceleration due to gravity = (7.86 × 10^3 kg) × (9.8 m/s^2)
Now, let's find the perpendicular distance for each sphere. The perpendicular distance is half of the diameter, which is equal to the radius.
For the iron sphere:
Radius(iron) = 1.40 cm = 0.014 m
Now, let's calculate the torque produced by each sphere. The torque is given by the formula: torque = force × perpendicular distance.
Torque(or force × radius) produced by the aluminum sphere = Force(aluminum) × Radius(aluminum)
Torque(or force × radius) produced by the iron sphere = Force(iron) × Radius(iron)
Since the torques should be equal to balance the spheres, we can equate the two torques:
Force(aluminum) × Radius(aluminum) = Force(iron) × Radius(iron)
Now, we can rearrange the equation and solve for the radius of the aluminum sphere:
Radius(aluminum) = (Force(iron) × Radius(iron)) / Force(aluminum)
Plug in the values we calculated before:
Radius(aluminum) = [(7.86 × 10^3 kg) × (0.014 m)] / [(2.70 × 10^3 kg) × (9.8 m/s^2)]
Now, calculate the radius(aluminum).