a car goes from a velocity zero to a velocity of 15 meters per second East in 2.1 seconds. What is the car's acceleration?
1 answer:
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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
<u>Answer:</u>
15.97 N force is tending to pull Rover forward
<u>Explanation:</u>
The woman pulls on the leash with a force of 20.0 N at an angle of 37° above the horizontal. The arrangement is shown in the given figure,
We nee to find the pulling force P. The 20.0 N force has two components, 20.0 cos 37 in horizontal direction and 20.0 sin 37 in vertical direction.
The horizontal component is equal to pulling force P, which will pull Rover forward/
So, P = 20.0 cos 37 = 15.97 N
15.97 N force is tending to pull Rover forward.
Answer:
Explanation:
For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:
where
q is the magnitude of the charge
v is the velocity
B is the magnetic field strength
In this problem,
So the force experienced by the electrons is