um ima em forma de barra atravessa uma espira retangular, como mostra a imagem. Sabendo que esse e um caso em que o fluxo magnet
ico sobre a espira esta variando, descreva o que acontecera na espira enquanto o ima se movimenta
1 answer:
Answer:De acuerdo con la ley de inducción electromagnética de Faraday, la tasa de cambio del flujo magnético vinculado con la bobina es directamente proporcional a la fem inducida en la bobina.
e = - d∅ / dt
Entonces, si el cambio de flujo es mayor, la fem inducida es mayor.
¡espero haber ayudado a tener un buen día! :)
Explanation:
According to the Faraday's law of electromagnetic induction, the rate of change of magnetic flux linked with the coil is directly proportional to the induced emf in the coil.
e = - d∅ / dt
So, if the flux change is more the induced emf is more.
i hope i helped have a good day! :)
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Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
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