An electron is moving with a velocity of 5 × 10 4 m/s (10 hold to the power of 4) enters into a uniform electric field and aquires a uniform acceleration of 10m/s2 in the direction of initial motion.
a) Calculate the time in which the electron would aquire a velocity double of its initial velocity.
b) How much distance would it cover in this time ?
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Sry nice try, but i realised it is wrong
Answered by
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Heya mate,
here is your answer,
_____________
Initial velocity of the electron is, u = 5 × 104 m/s
Acceleration of the electron is, a = 104 m/s2
(i)
Using, v = u + at
=> 2u = u + at
=> t = u/a = (5 × 104)/( 104) = 5 s
(ii)
Distance travelled in this time is,
S = ut + ½ at2
=> S = (5 × 104) × 5 + ½ × 104 × 52
=> S = 375000 m
_______________
# nikzz
HOPE U LIKE IT
CHEERS
☺☺
here is your answer,
_____________
Initial velocity of the electron is, u = 5 × 104 m/s
Acceleration of the electron is, a = 104 m/s2
(i)
Using, v = u + at
=> 2u = u + at
=> t = u/a = (5 × 104)/( 104) = 5 s
(ii)
Distance travelled in this time is,
S = ut + ½ at2
=> S = (5 × 104) × 5 + ½ × 104 × 52
=> S = 375000 m
_______________
# nikzz
HOPE U LIKE IT
CHEERS
☺☺
Thx
Sry nice try, but i realised it is wrong
The acc is not right check it ones in the question
i shall better not
only the 10 in the value of u is hold to the power of 4 not the acceleration one
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