A supertanker (mass = 1.57E+8 kg) is moving with a constant velocity. Its engines generate a forward thrust of 7.47E+5 N. Determine the magnitude of the resistive force exerted on the tanker by the water.
7.47×10^N is correct, but I need to know how
constant velocity (not accelerating), means the the thrust equals the resistive force
To determine the magnitude of the resistive force exerted on the supertanker by the water, we need to consider Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration.
In this case, since the supertanker is moving with a constant velocity, the net force acting on it is zero (assuming no other forces are acting on it).
The forward thrust generated by the engines of the supertanker is equal to 7.47 × 10^5 N. This forward thrust is balanced by the resistive force exerted by the water.
Therefore, the magnitude of the resistive force is also equal to 7.47 × 10^5 N.
To determine the magnitude of the resistive force exerted on the tanker by the water, you need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. However, since the supertanker is moving with a constant velocity (which implies a zero acceleration), the net force acting on it is zero.
In this case, the forward thrust provided by the tanker's engines (7.47×10^5 N) must be balanced by an equal and opposite resistive force exerted by the water. Therefore, the magnitude of the resistive force is 7.47×10^5 N.
To determine this, you must understand the concept of equilibrium. When an object is in equilibrium, the sum of all forces acting on it is zero. In this scenario, the forward force generated by the engines balances out the resistive force exerted by the water, resulting in no acceleration and a constant velocity.
M = 1.57*10^8 kg.
Ft = 7.47*10^5 N.
Fr = ?.
Ft-Fr = M*a.
Ft-Fr = M*0,
Ft-Fr = 0,
Fr = Ft = 7.47*10^5 N.