Answer:
B = - 1.51 10⁻⁷ T
Explanation:
For this exercise we can use Faraday's law of induction
E = -
In this case, they indicate that the normal and the magnetite field are in the same direction, so the angle is zero (cos 0 = 1), they give the area of the loop A = 4.32 10⁻⁴ m² and since we have N = 1470 turns in each one a voltage is induced
E = - N B A
B = -E A / N (1)
we find the induced voltage with ohm's law
V = i R
where the current is defined by
i = Q / t
we substitute
V = Q R / t
let's calculate
V = 9.18 10-3 56.0 / t
We must assume a time normally is t = 1 s
V = 0.514 V
this is the voltage in the circuit which must be the induced voltage V = E
we substitute in 1
B = - 0.514 4.32 10⁻⁴ / 1470
B = - 1.51 10⁻⁷ T
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Answer:
(c) 4M
Explanation:
The system is a loaded spring. The period of a loaded spring is given by

<em>m</em> is the mass and <em>k</em> is the spring constant.
It follows that, since <em>k</em> is constant,

where <em>C</em> represents a constant.


When the period is doubled,
.

Hence, the mass is replaced by 4M.
Answer:
2 0.00130719 m/s
Explanation:
Now we have
7.86 furlongs/fortnight
7.86 x 220 yards / fortnight
1729.2yards / fortnight
1647.8 yards / 14 days
123.5yards / day
So
123.5 x 0.9144 meter / day
112.meters / day
=112.9meters / 86400s
= 0.001307m/s