An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste ad it were already moving at speed v and the same amount of work W was done on it. What would be its final speed
1 answer:
Answer:
Explanation:
Given
W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)
According to work energy theorem work done by all the forces is equal to change in kinetic energy of object
where m=mass of object
v=velocity of object
When the object is already have velocity v then the final speed is given by work energy theorem
From 1 and 2 we get
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Electric Field = 3.369 x 10^4 N/C
Explanation:
Radius = r = (r1 + r2) / 2 = (1.6 + 3.6) /2 = 2.6 cm + 2.3 cm = 4.9 cm = 0.049 m
As we know, Electric field = E = kQ/r.r
= 8.98755 x 10^9 x 9 x 10^-9 / 0.049 x 0.049 = 33689.275 N/C
= 3.369 x 10^4 N/C
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We can use Newton's Second Law :
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We can rearrange the equation to solve for the acceleration.