A ball of mass 0.152 kg is dropped from a height 2.27 m above the ground. The acceleration of gravity is 9.8 m/s 2 . Neglecting
2 answers:
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The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i =
where R is the resistance
R =
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i =
q =
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
Answer:
I = 0.0025 kg.m²
Explanation:
Given that
m= 2 kg
Diameter ,d= 0.1 m
Radius ,
R=0.05 m
The moment of inertia of the cylinder about it's axis same as the disc and it is given as
Now by putting the all values
I = 0.0025 kg.m²
Therefore we can say that the moment of inertia of the cylinder will be 0.0025 kg.m².