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juin [17]
3 years ago
6

A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how

many turns are in the secondary? A. 8 B. 690 C. 72 D. 1,653
Physics
1 answer:
Nata [24]3 years ago
4 0

Answer:

C. 72

Explanation:

Transformer: A transformer is an electromagnetic device that uses the property of mutual inductance to change the voltage of alternating supply.

In a ideal transformer,

Vs/Vp = Ns/Np ............................................. Equation 1

Where Vp = primary voltage, Vs = secondary voltage, Ns = Secondary turn, Np = primary turn.

Making Ns the subject of the equation,

Ns =(Vs/Vp)Np .......................................... Equation 2

Given: Vs = 24 V, Vp = 115 V, Np = 345.

Substitute into equation 2

Ns = (24/115)345

Ns = 72 turns.

Thus the number of turns in the secondary = 72 turns.

The right option is C. 72

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Jlenok [28]

Answer:

-30°C

Explanation:

F-32/180 =C-0/100

or, -22-32/180=C/100

or, -54/180*100=C

or, -0.3*100=C

therefore, C= -30

-22°F = -30°C

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8 0
2 years ago
Where are you likely to find the oldest oceanic crust?
Stells [14]

Answer:

it would be letter E. near oceanic ridges

Explanation:

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4 0
3 years ago
You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
2 years ago
Help me with this review question please.
QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

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6 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
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