Answer:
B. 5 × 10¹ m³
Step-by-step explanation:
Let v represent the volume (in m³) of seawater added to the submarine's mass. Then the mass of that water is ...
v×10³ kg
After the water is added, the total mass of the submarine will be ...
(1.5×10⁵ + v×10³) kg
The density of the submarine is the ratio of this mass to its volume, so is ...
(1.5×10⁵ + v×10³) kg/(2×10² m³) = (750 +5v) kg/m³
We want this value to be 1000 kg/m³. Then we can find v to be ...
1000 = 750 +5v
250 = 5v
50 = v = 5×10¹ . . . . m³
The submarine needs to add 5×10¹ m³ of seawater to its tanks.
