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Dima020 [189]
2 years ago
7

The ratio of coils of wire in the primary winding to the secondary winding of a transformer is 3:2. How does the transformer cha

nge the voltage of a current applied to the primary winding?
It increases to three-halves times the original.
It decreases to two-thirds the original.
It increases to three times the original.
It decreases to one-third the original.
Physics
2 answers:
kykrilka [37]2 years ago
8 0

The correct answer is (B)... I am right!

LenaWriter [7]2 years ago
6 0
In a transformer, the ratio of the electromotive forces (EMFs), or voltages, at the primary and secondary ends of the transformer is equivalent to the ratio of the turns in the winding of the primary and secondary end. Mathematically:
E₁/E₂ = T₁/T₂
Given that T₁/T₂ = 3/2,
E₁/E₂ = 3/2
E₂ = 2/3 E₁
Therefore, the voltage decreases to two-thirds of the original value.
You might be interested in
Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time. Express
Inga [223]

Answer:

a=-\dfrac{k}{m}x(t)

Explanation:

From Newton's second law,

F = ma

where F is the force, m is the mass and a is the acceleration.

From Hooke's law,

F = -kx(t)

where k is the spring constant and x(t) is the displacement function measured from the origin. The negative sign indicates the force acts in opposite direction to the displacement. In fact, it is a restoring force; it acts to return the spring to its original undisturbed position.

Since both forces are the same,

F = ma= - kx(t)

a=-\dfrac{k}{m}x(t)

The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.

The ratio \dfrac{k}{m} is a constant except in situations where the mass is varying (say, the mass on the spring is a decaying material).

4 0
2 years ago
An electrical conductor is an element with __________ electrons in its outer orbit.
Setler [38]
An electric conductor is an element with free electrons in its outer orbit
5 0
2 years ago
A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t
arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vectorn=-0.60\hat{i}-0.080\hat{j}

We need to calculate the torque on the loop

Using formula of torque

\tau=NIAB\sin\theta

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}

\tau=4.7\times10^{-4}\ N-m

Hence, The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

5 0
2 years ago
Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD
NeX [460]

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

3 0
2 years ago
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
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