What are the STEPS to using a triple beam balance
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
Answer:
A. 2.83 m/s
B. 39.55 degrees north of east
Explanation:
The velocity of the ball relative to the ground is the sum vectors of velocity of the ball relative to Mia and the velocity of Mia relative to the ground
If we take north east as positive direction then
Velocity vector of the ball relative to Mia is
Velocity of Mia relative to gorund is
<0, 5.3>
So velocity of the ball relative to the ground is
<1.8, -3.12> + <0, 5.3> = <1.8, 2.18>
Its magnitude is:
m/s
Its direction is
north of east