"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after the pressure is applied."

Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

440 kg
view the diagram of the torques

let fg be the force of gravity exerted by the boat from its center of gravity
let x be the initial distance from the wheel pivot point.

rotational equilibrium so net torques ∑τ are equal to zero

we choose the pivot point to be the wheel so that we take advantage of both the 6m and the scale readings.

the scale readings mean that much mass is on it, so that means there is some gravity on it: mg = m(9.81m/s²)

before the boat is moved
∑τ = fg*x + (48 kg)(9.81 m/s²)(6 m) = 0 (i)

after the boat is moved
∑τ = fg*(x-0.15 m) + (37kg)(9.81 m/s²)(6m) = 0 (ii)

solve for x in (i)

fg * x + (48 kg)(9.81 m/s²)(6 m) = 0
fg * x  + 2825.28 nm = 0
x = -2825.28 nm / fg

substitute into (ii)

fg*(x-0.15m) + (37kg)(9.81 m/s²)(6m) = 0
fg*[ (-2825.28 nm / fg) - 0.15m] + 2177.82 nm = 0
-2825.28 nm - fg 0.15 m + 2177.82 nm = 0

-fg 0.15 m = 647.46 nm
fg = 4316 . 4 n

so the mass is

fg = mg
m = fg / g
m = (4316.4 n) / (9.81 m/s²)
m = 440 kg

440 kg