Help with a problem of a Sphere

[racmhead]

We have a hollow Sphere. It weighs 240 Newtons in the air, and 150 Newtons when it's totally immersed in the water.

If the density of the metal is 7 g/cm^3. Find the metal thickness

thanks

 

[Kurdt]

What have you tried so far?

 

[ogb p]

Based on the given information, we can use the formula for buoyancy force to solve for the volume of the sphere:

Buoyancy force = weight of displaced water

150 Newtons = (density of water) x (volume of sphere) x (acceleration due to gravity)

Solving for the volume of the sphere, we get:

Volume of sphere = 150 Newtons / (density of water) x (acceleration due to gravity)

= 150 Newtons / (1000 kg/m^3) x (9.8 m/s^2)

= 0.015 m^3

Since the sphere is hollow, we can use the formula for the volume of a hollow sphere to solve for the inner radius and outer radius:

Volume of hollow sphere = (4/3) x π x (outer radius^3 - inner radius^3)

0.015 m^3 = (4/3) x π x [(outer radius)^3 - (inner radius)^3]

Next, we can use the given weight of the sphere in air to solve for the outer radius:

Weight of sphere in air = (density of metal) x (volume of sphere) x (acceleration due to gravity)

240 Newtons = (7 g/cm^3) x (volume of sphere) x (9.8 m/s^2)

Solving for the volume of the sphere, we get:

Volume of sphere = 240 Newtons / (7 g/cm^3) x (9.8 m/s^2)

= 0.003 m^3

Substituting this into the previous equation, we get:

0.015 m^3 = (4/3) x π x [(outer radius)^3 - (inner radius)^3]

0.015 m^3 = (4/3) x π x [(outer radius)^3 - (outer radius - thickness)^3]

Simplifying and solving for the thickness, we get:

Thickness = (outer radius)^2 x (4π/3) x ((outer radius)^2 - 0.015 m^3) / (4π/3)

= (outer radius)^2 x ((outer radius)^2 - 0.015 m^3)

Using a calculator, we can solve for the outer radius to be approximately 0.093 m. Therefore, the thickness of the metal would be approximately 0.093 m. However, it is