Question

009 10.0 points

Answer 1

Answer:

Explanation:

Bulk modulus (B) =  2.3 x 10⁹ N/m²

B = Δp/ (Δv/v)

Δp = B (Δv/v)

density = mass / volume

d = m/ v

ln d = ln m - ln v

Δd/d = -Δv/v

Δp = B (Δd/d)

Δp = 360 - 1 = 359 atom

= 359 x 10⁵ N/m²

359 x 10⁵ =2.3 x 10⁹ x (Δd/d)

Δd = (359 x 10⁵/ 2.3 x 10⁹) x d

= 156.08 x 10⁻⁴ x 1045

= 16.31 kg/m³

increased density

= 1045 +16.31

= 1061.31 kg/m³

=

Answer 2

Answer:

The density of seawater at a depth is 1061.3 kg/m³.

Explanation:

Given that,

Pressure = 360 atm

Density at the surface = 1045 kg/m³

Bulk modulus $B=2.3\times10^{9}\ N/m$

We need to calculate the density of seawater at a depth

Using formula of density

$B=\rho_{0}\times\dfrac{\Delta P}{\Delta \rho}$

$\rho=\rho_{0}\times(1+\dfrac{\Delta P}{B})$

Put the value into the formula

$\rho=\rho_{0}\times(1+\dfrac{\Delta P}{B})$

$\rho=1045\times(1+\dfrac{(360-1)\times10^{5}}{2.3\times10^{9}})$

$\rho=1061.3\ kg/m^3$

Hence, The density of seawater at a depth is 1061.3 kg/m³.