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zheka24 [161]
3 years ago
8

**URGENT** Roberto plans to use two transformers to reduce a voltage of 120 V to 4 V. He uses a transformer that has 300 coils i

n the primary winding and 50 coils in the secondary winding. He has four other transformers, as listed in the table. Which transformer should Roberto use to have an ending voltage of 4 V?
A- W

B-X

C- Y

D- Z

Physics
2 answers:
Alexus [3.1K]3 years ago
9 0

Answer:

B) X

Explanation:

skelet666 [1.2K]3 years ago
3 0

As we know that in transformers we have

\frac{V_s}{V_p} = \frac{N_s}{N_p}

here we know that

V_s = 4 Volts

V_p = 120 Volts

N_s = 50 coils

N_p = 300 coils

now from above equation we will have

\frac{V}{120} = \frac{50}{300}

V = 20 Volts

now we have to reduce this voltage to final voltage of V = 4 V

so again we will have

\frac{V_s}{V_p} = \frac{N_s}{N_p}

\frac{4}{20} = \frac{N_s}{N_p}

\frac{N_s}{N_p} = \frac{1}{5}

so we need to take such a winding whose ratio is 1:5

So it is satisfied in X

N_p = 60

N_s = 12

so answer will be

<u>B)-   X</u>

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4 0
3 years ago
Why can an object still be seen when it is at absolute zero?
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6 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
Given that: = 2i + 9j +3k and = -i – 4k . Find
Igoryamba

Answer:

3i+9j= -7k

Explanation:

2i+I+9j=-4k-3k

3i+9j= -7k

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What is meant by the term "observable universe"
Marta_Voda [28]

Answer:

The observable universe is a ball-shaped region of the universe comprising all matter that can be observed from Earth or its space-based telescopes and exploratory probes at the present time

Explanation:

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