(a) what Is the density of an object that is 14% submerged when floating in water at 0 degrees Celsius? (b) what percentage of the object will be submerged if it is placed in ethanol at 0 degrees Celsius?
d1*V1 = d2*V2
where d1 and V1 are the density and volume of object 1; d1 and V2 are the density and volume of the fluid
Then V2 = 0.14*V1
d1*V1 = d2*0.14*V1
d1 = d2*0.14
Look up the density of water at 0 degrees C (= d2). Solve for d1
b) d1*V1 = d2*V2
You know d1 from a. You can look up the the density of ethanol, d2.
% submerged = V2/V1 = d1/d2
To determine the density of the object in question, we need to use a formula involving the submerged fraction, which is the ratio of the volume submerged to the total volume of the object.
(a) To calculate the density of the object when it is 14% submerged in water:
1. Firstly, let's assume the density of water at 0 degrees Celsius is ρ_water = 1000 kg/m³.
2. The submerged fraction is given as 14%, which can be written as 0.14.
3. The buoyant force acting on the object is equal to the weight of the water it displaces (Archimedes' principle).
4. As the object is floating, the buoyant force is equal to the weight of the object itself.
5. Let's assume the volume of the object is V_object.
6. The volume submerged will be 0.14 * V_object.
7. The buoyant force is then (0.14 * V_object) * ρ_water * g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
8. The weight of the object is equal to its mass multiplied by the acceleration due to gravity, which is W_object = m_object * g.
9. Since the object is floating, the weight of the object is balanced by the buoyant force, so we have:
(0.14 * V_object) * ρ_water * g = m_object * g
Simplifying, we get:
0.14 * ρ_water = ρ_object
Therefore, the density of the object is approximately 0.14 times the density of water at 0 degrees Celsius.
(b) To determine the percentage of the object that will be submerged in ethanol at 0 degrees Celsius, we need to repeat the same steps as in part (a), but with the density of ethanol instead of water.
1. Assume the density of ethanol at 0 degrees Celsius is ρ_ethanol (lookup).
2. Follow steps 2-9 from part (a) with ρ_ethanol replacing ρ_water.
3. Calculate the resulting value for ρ_object.
4. To find the submerged fraction in ethanol, divide ρ_object by ρ_ethanol.
5. Multiply by 100 to get the percentage.
Please note that the densities of different liquids can vary, so it is important to accurately determine the density of the specific liquid you are using in your calculations.