Question

A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300 meters long, 80 meters wide, and 40 meters high, determine how far the bottom of the ship is below sea level. 4. [psea 1 020 kgm3 =
a) 10 m
b) 15 m
c) 20 m
d) 25 m
e) 30 m

Answer 1

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × $10^{8} \ kg$

$\rho_{sea} = 1020\ \frac{kg}{m^{3} }$

We know that Buoyant force on the tank is equal to gravity force of the tank.

$F_B = F_g$

$(\rho_{Fluid}) (g) (V_{disp}) = m g$

$(\rho_{Fluid}) (V_{disp}) = m$

1020 × $V_{disp}$ = 6.1 × $10^{8}$

$V_{disp}$ = 598039.21 $m^{3}$

We know that

$V_{disp}$ = W × L × H

598039.21 = 300 × 80 × H

H = 25 m

Therefore the bottom of the sea is 25 m below sea level.