Ask question

Ask question

All categories

r-ruslan [8.4K]

2 years ago

5

The voltage across the secondary winding is 10,000 V, and the voltage across the primary winding is 25,000 V. If the primary win

ding has 200 coils, how many coils are in the secondary winding?

2 answers:

arlik [135]2 years ago

8 0

Answer:

The answers is 80 Coils

Explanation:

Right on edge

Allisa [31]2 years ago

7 0

<u>Answer</u>

80 coils

<u>Explanation</u>

The turn rule of a transformer says that, the amount of voltage induced in the secondary coil is proportional to the number of coil. It is simplified by the ratio:

N<em>p/</em><em>N</em><em>s = </em><em>V</em><em>p/</em><em>V</em><em>s</em>

200/Ns = 25,000/10,000

200 × 10,000 = 25,000Ns

2,000,000 = 25000Ns

Dividing both sides by 25,000,

Ns = 2,000,000/25,000

= 80 coils

You might be interested in

A submarine is at an ocean depth of 3.00 km.

Alexxx [7]

Answer:

1.#(#('#;#+#('#;#+jwhsjjshajahkqhwmskek

Explanation:

jsjsnanakamakwjKbakqwigjgodrgoshoihrdohrf

tjfjrhgcorfhptgkc

jv

tgj

crlfkhc

rjv

tkgjv

kyjv

tgjvhrlkrfjhirghogrtiguotightjptrjpghpf

rofrirhfohrihithfitgprirl

ceohr

fbojofjigoitigotjog

irojfofjkrhrjoejoqjodirnkfnkrjoritpruixhor

firhofeiidiehdijeihr
jdhxjdhjchfihvidjfdoigorjgiifugiruogu
ifgijrojiru

ifhfojrorior

bifjibjtgitkriwgi eror iurh oryi ehiwyrfoerhfoefoeuis in het begin van een tui ueen uitgebreide beschrijving ygt grup u een uitgebreide keuze u guugu

sjksjsosjsnshf ruor uw eigen website aanmaken r y fav u can u to uf the owner u will tu tutururu ba sa inyo ay walang naging y rin h results ry from y yifuirfjofjkrjirhieifr Fu t I

7 0

2 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

2 years ago

Read 2 more answers

The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a

diamong [38]

Answer: $\theta =9.96^{\circ}$ North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus $195sin\theta =33.9$

$\theta =9.96^{\circ}$

so Plane should head towards $9.96^{\circ}$ North of west in order to fly in west.

So plane

actual velocity is

$v=-195cos9.96\hat{i}+195sin9.96\hat{j}$

5 0

2 years ago

If a 5.00 kg box slides down a ramp inclined at 60.0° above the horizontal, what is the

Basile [38]

Answer:

A 70 kg box is slid along the floor by a 400 n force. The coefficient of friction between the box and the floor is 0. 50 when the box is sliding

8 0

2 years ago

The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

3 years ago