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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Explanation:
Matter is changed from one state to another by addition or removal of heat and suitable pressure conditions.
When a solid is heated, it normally melts and changes to liquids which on heating changes to vapor. The randomness of the particles increases from solid to liquid state and to gaseous states.
Also, a gas can be cooled to liquid and on further cooling transformed into a solid matter.
These phase changes are brought about by energy changes in a system. Some form of matter can also sublime by changing form solid to gas and vice versa.
Answer:
KE = KE (incidental) - KE of emitted photons
or KE = h * f - Wf
So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy
If you double the frequency then h * f = 6.16
KE = 6.16 - 1.2 = 4.96 eV
