a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W
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Answer: The motion of the object will remain the same
Explanation:
Answer:
8 N North.
Explanation:
Given that,
One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.
We need to find the magnitude of net force acting on the object.
Let North is positive and South is negative.
Net force,
F = 10 N +(-2 N)
= 8 N
So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".
Explanation:
Given that,
The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.
(a) Capacitance of capacitor 1,
Capacitance of capacitor 2,
Capacitance of capacitor 3,
(b) The equivalent capacitance in series combination is :
Hence, this is the required solution.
