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Alborosie
3 years ago
14

A transformer is to be used to step down voltage from an alternating current source from 220V to 110V. If the primary has 120 tu

rns, then the number of turns in the secondary is.
Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

The number of turns on the secondary would be 60.

Explanation:

The voltage is to be cut in half. The ratio between the primary and secondary, for a step-down transformer, would then be 2:1.

120 turns / 2 = 60 turns

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How many electrons must be remowel from an electricaly Nurutral Silvor Coin to five it a charge of 3.2 NC ?
kakasveta [241]

Answer:

 #_electrons = 2 10¹⁰ electrons

Explanation:

For this exercise we can use a direct rule of three proportions rule. If an electron has a charge of 1.6 10⁻¹⁹ C how many electrons have a charge of 3.2 10⁻⁹ C

          #_electrons = 3.2 10⁻⁹ ( \frac{1}{1.6 \ 10^{-19}})

          #_electrons = 2 10¹⁰ electrons

7 0
3 years ago
HELP ME PLEASE
Stels [109]
The answer is B. the germinal stage
7 0
3 years ago
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Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
victus00 [196]

Electric field lines always begin at positive charges (or at infinity) and end at negative charges (or at infinity).

One could also say that the lines we use to represent an electric field indicate the direction in which a positive test charge would initially move when released from rest.

6 0
4 years ago
Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo
klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J      

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy (E_{k}) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

                                        E_{k} = h(f − f₀)

                                        E_{k} = hf - hf₀

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

                                        E_{k} = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy E_{k} = 4.16×10⁻¹⁷ J  

Speed of light  c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js                                

                                        E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

                                        E = 53.8356 x 10⁻¹⁶ J

from E_{k} = E - ϕ ;

                                        ϕ = E - E_{k}

                                        ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

                                        ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J      

4 0
3 years ago
Since acid solutions are conductors of electricity, they can be used in batteries. true false
Eduardwww [97]
This is true, I believe. 

Hope this helps!
8 0
4 years ago
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