The pressure at a depth of 300 m in the ocean exceeds sea-level atmospheric pressure by 3.03 kPa. By how much does the volume of a 0.2-m3 aluminum object contract when lowered to this depth in the ocean? The bulk modulus of aluminum is 7 x 10^10 Pa.
To find the change in volume of the aluminum object when lowered to a depth of 300 m in the ocean, we will use the concept of bulk modulus. The formula to calculate the change in volume is given by:
ΔV = -V * (ΔP / B)
Where:
ΔV is the change in volume
V is the initial volume of the object
ΔP is the change in pressure
B is the bulk modulus of the material
In this case, the initial volume of the aluminum object, V, is 0.2 m³.
The change in pressure, ΔP, is the difference between the pressure at a depth of 300m and the sea-level atmospheric pressure, which is 3.03 kPa.
Now, let's calculate the change in volume:
ΔV = -V * (ΔP / B)
= -0.2 m³ * (3.03 kPa / (7 x 10^10 Pa))
First, we need to convert the pressure from kilopascals (kPa) to pascals (Pa):
ΔP = 3.03 kPa * 1000 Pa/kPa
= 3030 Pa
Now, let's substitute the values and calculate the change in volume:
ΔV = -0.2 m³ * (3030 Pa / (7 x 10^10 Pa))
≈ -0.00008657 m³
Therefore, the volume of the aluminum object contracts by approximately 8.657 x 10^-5 m³ (cubic meters) when lowered to a depth of 300 m in the ocean.