Answer:
1215 N
Step-by-step explanation:
We're told from the question that the drag force, F, on a boat varies jointly with the wet surface area, A, of the boat, and the square of the speed, s, of the boat, this can be expressed mathematically as;

where k = constant of proportionality
Given A = 83 ft^2
s = 20 mph
F = 996N
Let's go ahead and substitute these values into the above equation and solve for k;

Given A = 125 ft^2, s = 18 mph and we know that k = 0.03, let's go ahead and solve for F;
