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tangare [24]
3 years ago
11

the very high voltage needed to create a spark across the spark plug is produced at the a. transformer's primary winding. b. tra

nsformer's secondary winding. c. battery. d. breaker points.
Physics
2 answers:
Karo-lina-s [1.5K]3 years ago
5 0
I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the  transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>
Luden [163]3 years ago
4 0

Answer:

b ) transformer's secondary winding.

Explanation:

When we switch on the car , the breaker point is closed sending current through primary winding of a transformer creating magnetic field in it. When the point is  opened again  , current stops in the primary coil ,reducing magnetic field to zero. This creates very high voltage in the secondary winding . 12 volt of car battery is transformed into 30000 volt in secondary coil. This high voltage is carried to spark plug by high tension wire.

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A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
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Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

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total distance = 1868.478 m

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