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OLEGan [10]
3 years ago
12

If the primary of a transformer were connected to a dc power source,

Physics
1 answer:
QveST [7]3 years ago
8 0

Answer:

D. only briefly while being connected or disconnected.

Explanation:

As we know that transformer works on the principle of mutual inductance

here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil

So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.

Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always

So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output

so correct answer will be

D. only briefly while being connected or disconnected.

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3 years ago
An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

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3 years ago
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up
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Answer:

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F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

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