Find the Apparent Weight: Solving for Volume

[akatsafa]

An object hanging in the air causes a scale to read 0.0451kg. When the object was submerged in water, the scale is balanced at 0.0370. I found the object of the weight in air to be 0.442N. The question is to find the apparent weight when submerged in water. How do I find the volume to find the apparent weight. So far, I have 0.442-(1000kg/m^3)(9.8m/s^2)(V)=W. What's the volume?

 

[Doc Al]

I presume that by "apparent weight" is meant what the scale reads when the object is submerged. You measured the apparent weight directly, so what are you calculating?

 

[M98Ranger]

To find the apparent weight when submerged in water, you will need to use the buoyant force equation, which is given by Fb = ρVg, where Fb is the buoyant force, ρ is the density of the fluid (in this case, water), V is the volume of the object, and g is the acceleration due to gravity.

In this case, the buoyant force is equal to the weight of the water displaced by the object, which is also known as the apparent weight. So, to find the apparent weight, you will need to rearrange the equation to solve for V.

V = (W - Fb)/(ρg)

Substituting the values given in the problem, we get:

V = (0.442N - 0.037N)/(1000kg/m^3)(9.8m/s^2)

V = 0.405m^3

Therefore, the volume of the object is 0.405m^3. This means that when the object is submerged in water, the apparent weight is equal to the weight of the water displaced, which is 0.405m^3 x 1000kg/m^3 x 9.8m/s^2 = 3.96N.

To summarize, to find the apparent weight when submerged in water, you need to use the buoyant force equation and solve for V. Then, you can calculate the apparent weight by multiplying the volume by the density of the fluid and the acceleration due to gravity.