The electron relaxation time is
The mean free path is
<u>Explanation:</u>
Electron density,
Aluminium resistivity,
From the Drude's model we have:
Where:
τ= Electron relaxation time
m = mass of a charge
q = magnitude of a charge
We know, electron mass =
Charge of electron =
By substituting all given values for electron, we get
When multiply by 100 and divide by 100, we get
Mean free path is given as:
where:
l = Mean free path
= Average velocity of electrons
We know the general value for average velocity of electrons at room temperature:
Therefore,
Answer:
The objects were 1.8m apart.
Explanation:
We will start stating the Coulomb's Law. It says that:
Where F_e is the electric force between the objects, q_1 and q_2 are the magnitude of the charge of the objects, r is the distance between them and K is the Coulomb's constant ( in vacuum). Solving for the distance r we have:
Plugging the given values into this equation, we obtain:
In words, the two charged objects were 1.8m apart.
Force = mass x acceleration
113N = 25.2a
a = 4.48 m/s^2
Vf = Vi + at
Vf = 2.33 + 4.48(2) = 11.3 m/s
Answer:
8.9*10^6 V/m
Explanation:
The expression for electric field strength E is given as
where V= voltage
d= distance of separation
Given data
substituting our given data into the electric field strength formula we have
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.