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DENIUS [597]
3 years ago
11

A constant voltage of 3.00 V has been observed over a certain time interval across a 3.00 H inductor. The current through the in

ductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval?
Physics
1 answer:
frozen [14]3 years ago
5 0

Answer:

Explanation:

If L be inductance and i be current , flux associated with it

Φ = L i

emf induced = dΦ / dt = L di / dt

Putting the given values

emf induced = 3 V , L = 3 H

di = 8 - 3 = 5 A

dt = ?

3 = 3 x 5 / dt

dt or time duration = 3 x 5 / 3

= 5 sec .

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DedPeter [7]

Answer:

<u>Example of Newton's III law</u>

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8 0
3 years ago
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of gla
Diano4ka-milaya [45]

Complete Question

A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,  

A :

a force repels the glass out of the capacitor.  

B :

a force attracts the glass into the capacitor.    

C :

no force acts on the glass.      

D :

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E :

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Answer:

The correct option is B

Explanation:

Generally when the glass dielectric is slowly inserted between the plated,

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3 0
3 years ago
MATCH THESE ^-^ Match Newton's law with the correct statement.
ikadub [295]
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4 0
3 years ago
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Murrr4er [49]

Answer:

event s

Explanation:

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8 0
3 years ago
Pa help po science po yan pang grade 7
AveGali [126]

Answer:

table 1:

1. 100/15 = 6.7 m/s

2. 100/12 = 8.3 m/s

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Explanation:

8 0
3 years ago
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