Review question please take a look and help
1 answer:
You might be interested in
Answer:
(a) 1.87 x 10⁻⁴ m/s
(b) 0.013V
Explanation:
(a) Drift speed, , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;
=
----------------(i)
Where;
q = amount of charge
n = free charge density
A = cross-sectional area of the wire
But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e
J = =
Equation (i) can then be written as follows;
=
=
=
---------------------(ii)
From the question;
E = 0.0520N/C
p = 1.72 x 10⁻⁸ Ωm
n = 8.5 x 10²⁸ electrons/m³
c = charge on electron = 1.9 x 10⁻¹⁹C
Substitute these values into equation (ii) as follows;
=
= 1.87 x 10⁻⁴ m/s
(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e
V = E x d [where d = 25.0cm = 0.25m]
V = 0.0520 x 0.25
V = 0.013V
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
Substituting into the equation, we have;
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g