Question

if an electron moving with aspeed of 2.5x10^7 ms^-1 is deflected by an electric field of 1.6kVm^-1 perpendicular to its circular path then c/m for the electron will be (given radius of circular path=2.3m) a.1.7x10^11ckg^-1 b.1.8x10^11ckg^-1 c.1.9x10^-19 d.1.85x10^11 Ckg^-1

Answer 1

Answer:

hey mate

Explanation:

Given in the question ,

Here mass of the block = 100 g = 0.1 kg 

Velocity v = 5 m/s

Now for Initial Kinetic Energy .

= 1/2 mv² 

= 1/2 (0.100 x 5²)

= 0.5 x 2.5

= 1.25 J.

Now change in Kinetic energy = 0 - 1.25   {Final K.,E = 0]

= -1.25

Therefore diff B/w it's original position & final position =Diameter of tube. 

h= 2 x 10 m

h= 0.20 m

Thus work done by the gravity is 

= mgh

= 01 x 9.8 x 0.2

W = 0.2 J

Here, change in Kinetic energy = work done by gravity - work done by tube.

Thus, Work done by tube = W.D by gravity - Change in Kinetic energy

= 0.2 - (-1.25)

= 0.2 +1.25

= 1.45 J.

Here the movement will always against the force since the W.D is negative.

= -1.45

So the work done by the tube on the block is -1.45 J.

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