Answer:
I = 5[amp]
Explanation:
Electrical power is defined as the product of voltage by current.
where:
P = power = 1150 [W]
V = voltage = 230 [V]
I = current [amp]
Now replacing:
A 15 [amp] fuse must be used. Always the fuse must be larger than the operating current, to protect the equipment from very high currents. above 15 [amp]
Answer:
F = 3.01 N
Explanation:
Given that,
Mass of a block, m = 2.75 kg
Force applied to the block, F = 5.11 N
It is directed 53.8° above horizontal.
We need to find the total force acting on the block. The force acting on it is given by :
So, 3.01 N of force is acting on the block.
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