A hollow metal sphere of mass 5kg is tied to the bottom of the sea-bed by a rope. The tension or force in the rope is 60 N.

Calculate the volume of the sphere.?

the force is equal to the weight of the water displaced.

Divide that weight (-5*9.8 N) by density of seawater, and that is the volume of the sphere. Watch units.

Physics question.

Chemistry question.

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To calculate the volume of the hollow metal sphere, we can use the principle of buoyancy.

The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the fluid is water, and the buoyant force is equal to the tension or force in the rope.

Given that the tension in the rope is 60 N, which is equal to the buoyant force, we can use the equation:

Buoyant force = Weight of water displaced

The weight of water displaced can be calculated using the equation:

Weight = mass x gravitational acceleration

The mass of the water displaced is equal to the mass of the sphere because it is completely submerged in water. The gravitational acceleration (g) is approximately 9.8 m/s^2.

So we have:

Buoyant force = Weight of water displaced
60 N = mass of sphere x gravity

Rearranging the equation, we can solve for the mass of the sphere:

mass of sphere = Buoyant force / gravity
mass of sphere = 60 N / 9.8 m/s^2
mass of sphere ≈ 6.12 kg

Now that we have the mass of the sphere, we can use the formula for the volume of a sphere to calculate its volume:

Volume = (4/3) x π x radius^3

Since the sphere is hollow, we need to subtract the volume of the inner cavity from the total volume. Let's assume the outer radius of the sphere is R and the inner radius is r.

Let's substitute the values into the formula:

5 kg = (4/3) x π x R^3 - (4/3) x π x r^3

Now we need to solve for R. Rearranging the equation:

5 kg = (4/3) x π x R^3 - (4/3) x π x r^3
5 kg = (4/3) x π (R^3 - r^3)

To solve for R, we need the value of r. Unfortunately, the equation doesn't provide information about the inner radius of the hollow sphere. Without this information, it is not possible to calculate the volume of the sphere.