Physics: 2 pts.) An electron is placed in an electric field of intensity 700 N/Cj. What are themagnitude and direction of the acceleration of this electron due to this field? (melectron=9.1 × 10-31 kg, qe= 1.60 × 10-19 C)

2 pts.) An electron is placed in an electric field

babykate66

28.01.2022

Explanation:

We have,

Electric field intensity is 700 N/Cj

It is required to find the magnitude and direction of the acceleration of this electron due to this field.

The electric force is balanced by the force due to its motion i.e.

qE =ma

a is acceleration of this electron

$a=\dfrac{qE}{m}\\\\a=\dfrac{1.6\times 10^{-19}\times 700}{9.1\times 10^{-31}}\\\\a=1.23\times 10^{14}\ m/s^2$

So, the acceleration of the electron is $1.23\times 10^{14}\ m/s^2$ and it is moving in +y direction.

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alsiedlaw

16.05.2023

Explanation:

The question says that, "After how long does a mobile acquire a speed of 30 m / s, if it starts from rest and its acceleration is 3 m / s*2"

Initial speed is 0

Final speed is 30 m/s

Acceleration a = 3 m/s²

It is required to find time when it acquire a speed of 30 m/s. Using equation of kinematics as :

$v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{30}{3}\\\\t=10\ s$

So, after 10 seconds it will acquire a speed of 30 m/s.

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