The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
<h3>
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
<span>Lets call F the friction force which will act horizontally backwards.
As you are travelling at a constant velosity horizontally there is no overall resultant force in this direction.
ie. the force you pull with will be equal to the friction force resisting you. (you will initially have to have pulled with a greater force than the friction to get the suitcase moving)
the value of your force pulling is 60 cos26.9 (horizontally) - you should have learnt about resolving forces.
this must be equal to F
so
F=60cos26.9
F=53.5N
hope this helps you
please mark this as brainliest answer</span>
Answer:
240cm
Explanation:
Speed = 6cm/s
Time = 40 seconds
Speed= distance/time
6cm/s = distance/40seconds
Distance = 6×40
Distance = 240cm
Hence, in 40 seconds, the ball must have rolled the distance of 240cm
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v
charge on 2.0μf capacitor is
=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
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