Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
Mobility
We can find the drift velocity replacing,
The electric field is given by,
Option a; Electric field can accelerate an electron, but never change its speed
An electric field (also known as an E-field) is a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It can also refer to the physical field of a charged particle system. Electric fields are created by electric charges and time-varying electric currents. Electric and magnetic fields are both aspects of the electromagnetic field, one of nature's four fundamental interactions (also known as forces). Electric fields are significant in many areas of physics and are used in electrical technology. In atomic physics and chemistry, for example, the electric field is the attractive force that holds the atomic nucleus and electrons together in atoms. It is also the driving force behind chemical bonds between atoms.
Learn more about Electric field here:
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The rocket will cover 
distance in 4. 5 s. Acceleration can be defined as the change in velocity.
<h3>
What is acceleration?</h3>
Acceleration can be defined as the change in speed or the direction of the object.
From kinamatic equation:
Where,
- final velocity = 445 m/s
- initial valocity = 0 m/s
- acceleration = 99. 0 m/s²
- time = 4. 50 s
Put the values in the formula,
Therefore, the rocket will cover distance in 4. 5 s.
Learn more about Acceleration :
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Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
