The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
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Work done by the force experienced by the object</h3>
The magnitude of the work done by force experience by the object is calculated as follows;
W = f.d
where;
- F is the applied force (2xyi + 3yj), where x and y are in meters
- d is the displacement of the object = (a, b)
The work done by the force is determined from the dot product of the force and the displacement of the object.
F = (2xyi + 3yj).(a + b)
W = (2abi + 3bj).(ai + bj)
W = (2a²b + 3b²)J
Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
The complete question is below:
The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.
How much work does the force do?
Learn more about work done here: brainly.com/question/8119756
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Answer:
T = 1.766(M-m) Nm where M and m are the 2 masses of the objects
Explanation:
Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.
Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m
T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm
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