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
Answer:
the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Explanation:
This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Wavelength is defined as distance between two troughs or two crests !!
so in this wavelength is 4 metres !!
Answer:
atom(which contains protons, neutrons and electrons)
Answer:
Explanation:
It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.
