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Inessa05 [86]
3 years ago
8

A step-down transformer has more loops on which coil?

Physics
2 answers:
Nesterboy [21]3 years ago
6 0
A step-down transformer has more loops in :  A. Primary coil

Primary coil refers to the coil to which alternating voltage is supplied. It's usually connected to the AC supply

hope this helps
MrRissso [65]3 years ago
3 0

Answer:

A). Primary coil

Explanation:

As we know that in an ideal transformer the flux linked with primary coil and secondary coil for each turn must be same

so we can say that

\phi _p = N_p \phi_0

\phi_s = N_s\phi_0

so as per faraday's law we can say that two EMF is given as

E_s = N_s\frac{d\phi_0}{dt}

E_p = N_pfrac{d\phi_0}{dt}

now we have

\frac{E_s}{E_p} = \frac{N_s}{N_p}

so we can say that

for step down transformer the output voltage is less than the input transformer

E_s < E_p

so we will have

N_s < N_p

so turns will be more in

A) Primary coil

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Lady_Fox [76]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

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\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}

3 0
2 years ago
A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch
ololo11 [35]

Answer:

 v_{1f} = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

      p₀ = (m₁ + m₂) v₀

After the explosion

      pf = m₁ v_{1f} + m₂ v_{2f}

     p₀ = [texpv_{f}[/tex]

     (m₁ + m₂) v₀ = m₁ v_{1f} + m₂ v_{2f}

Let's calculate the final speed (v1f) of the first stage

     v_{1f} = ((m₁ + m₂) v₀ - m₂ v_{2f}) / m₁

     

     v_{1f} = ((2100 +1160) 4300 - 1160 5940) / 2100

     v_{1f} = (14,018 10 6 - 6,890 106) / 2100

     v_{1f} = 7,128 106/2100

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come the same direction of the final stage, but more slowly

4 0
3 years ago
Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the
murzikaleks [220]

Answer:

The function has a maximum in x=3

The maximum is:

f(3) = 39

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

f(x)' = -4*2x + 24=0

-4*2x + 24=0

8x=24

x=3

Now find the second derivative of the function and evaluate at x = 3.

If f (3) ''< 0 the function has a maximum

If f (3) '' >0 the function has a minimum

f(x)''= 8

Note that:

f(3)''= -8

the function has a maximum in x=3

The maximum is:

f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39

4 0
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yKpoI14uk [10]

Answer:

11760 joules

Explanation:

Given

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Height (h) = 16m

Required

Determine the increment in potential energy (PE)

This is calculated as thus:

PE = mgh

Where g = 9.8m/s²

Substitute values for m, g and h.

P.E = 75 * 9.8 * 16

P.E = 11760 joules

8 0
3 years ago
A rock is thrown horizontally from a bridge with a speed of 29.0 m/s. if the rock is 23.7 meters above the river at the moment o
mash [69]
It would be 1.5 meters im sure form that distance to me is that nswe

7 0
2 years ago
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