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Answer :
Answer:
The magnitude of the force from the wind is approximately 65,370 N.
Explanation:
The work done by the wind on the boat is given by the equation:
W = F * d * cos(theta)
where W is the work done (4.950 MJ), F is the force from the wind, d is the distance traveled (258 m), and theta is the angle between the direction of the force and the direction of travel (74 degrees).
Rearranging the equation to solve for F, we get:
F = W / (d * cos(theta))
Substituting the given values, we get:
F = (4.950 * 10^6 J) / (258 m * cos(74 degrees))
Using a calculator, we find that cos(74 degrees) is approximately 0.2756, so:
F = (4.950 * 10^6 J) / (258 m * 0.2756)
F = 65,370 N
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Rewritten by : Barada
To find the magnitude of the force from the wind, we need to use the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy:
work = ΔK
Here, the work done by the wind is 4.950 MJ, or 4.950 × 10^6 J. The boat's initial velocity is zero, and its final velocity is unknown. However, we know that the boat travels a distance of 258 m [N], so we can use this information to find its final velocity using the equation:
d = (1/2) (v_i + v_f) t
where d is the distance, t is the time, and v_i and v_f are the initial and final velocities, respectively.
Since the boat is traveling directly north, we can assume that the force from the wind is perpendicular to its path, so it does not affect its speed in the north direction. Therefore, we only need to consider the boat's motion in the east direction.
Using the work-energy principle and the equation above, we can write:
work = ΔK
4.950 × 10^6 J = (1/2) m (v_f^2 - 0)
where m is the mass of the boat, and v_f is its final velocity in the east direction.
Solving for v_f, we get:
v_f = sqrt((2 work) / m)
v_f = sqrt((2 * 4.950 × 10^6) / 182)
v_f = 247.8 m/s
The boat's final velocity is 247.8 m/s [E74°N]. To find the magnitude of the force from the wind, we can use the equation:
work = Fd cos θ
where F is the magnitude of the force, d is the distance traveled, and θ is the angle between the force and the direction of motion.
In this case, θ = 74°, and d = 258 m. Substituting these values and solving for F, we get:
F = work / (d cos θ)
F = 4.950 × 10^6 / (258 cos 74°)
F = 38,744 N
Therefore, the magnitude of the force from the wind is 38,744 N.
work = ΔK
Here, the work done by the wind is 4.950 MJ, or 4.950 × 10^6 J. The boat's initial velocity is zero, and its final velocity is unknown. However, we know that the boat travels a distance of 258 m [N], so we can use this information to find its final velocity using the equation:
d = (1/2) (v_i + v_f) t
where d is the distance, t is the time, and v_i and v_f are the initial and final velocities, respectively.
Since the boat is traveling directly north, we can assume that the force from the wind is perpendicular to its path, so it does not affect its speed in the north direction. Therefore, we only need to consider the boat's motion in the east direction.
Using the work-energy principle and the equation above, we can write:
work = ΔK
4.950 × 10^6 J = (1/2) m (v_f^2 - 0)
where m is the mass of the boat, and v_f is its final velocity in the east direction.
Solving for v_f, we get:
v_f = sqrt((2 work) / m)
v_f = sqrt((2 * 4.950 × 10^6) / 182)
v_f = 247.8 m/s
The boat's final velocity is 247.8 m/s [E74°N]. To find the magnitude of the force from the wind, we can use the equation:
work = Fd cos θ
where F is the magnitude of the force, d is the distance traveled, and θ is the angle between the force and the direction of motion.
In this case, θ = 74°, and d = 258 m. Substituting these values and solving for F, we get:
F = work / (d cos θ)
F = 4.950 × 10^6 / (258 cos 74°)
F = 38,744 N
Therefore, the magnitude of the force from the wind is 38,744 N.
