Answer to Question #265784 in Electric Circuits for Jayson

Question

The deepest point in the ocean is in the Mariana Trench, about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 x 10^8 Pa. (a) Calculate the change in volume of 1.00 m3 of seawater carried from the surface to this deepest point. (b) The density of seawater at the surface is 1.03 x 10^3 kg/m3. Find its density at the bottom. (c) Explain whether or when it is a good approximation to think of water as incompressible.l

Answer 1

Part(a)

d=11km=11000m

$\rho=1.03\times10^3kg/m^3$

$p_1=1.13\times10^8 Pa\\p_2=1.01\times10^5Pa$

$$∆P=1.13\times10^{8}-1.01\times10^{5}=1.12899\times10^8Pa$$

$\beta=-V\frac{∆p}{∆V}$

$2.1\times10^{9}=-1\times\frac{1.12899\times10^8}{∆V}$

$∆V=-0.05377m^3$

Part(b)

$$∆\rho=\frac{m∆\rho}{V}=\frac{0.053\times10^3}{1}=0.053\times10^3kg/m^3$$

$∆\rho=\rho_2-\rho_1$

$\rho_2=∆\rho+\rho_1$

$$\rho_2=(0.053+1.03)\times10^3kg/m^3=1.08\times10^3kg/m^3$$

Part(c)

Yes it is a good approximation because in the 11km the density of water changed by only (5-6)%.