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2 years ago

Define refractive index.

2 answers:

8 0

The refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.

the ratio of the velocity of light in a vacuum to its velocity in a specified medium.

irga5000 [103]2 years ago

5 0

Answer:

In optics, the refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.

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A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

6 0

3 years ago

How many neutrons does element x have if it's atomic number is 27 and it's mass number is 74

Makovka662 [10]

Atomic number=Proton count

Atomic mass=Proton count+ neutron count

Neuton Count=Atomic mass-Proton count

Proton count=Atomic number=27

Mass number=74

Neuton count= 74-27=47

3 0

3 years ago

Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the poten

alukav5142 [94]

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

$V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}$

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

$V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}$

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

4 0

3 years ago

6th grade science I mark as brainliest.

DerKrebs [107]

Answer:

2m 13 $\frac{1}{3}$ s

Explanation:

1.5m = 1s

200m = $\frac{200}{1.5}$ × 1s

= 133 $\frac{1}{3}$ s

= 2m 13 $\frac{1}{3}$ s

3 0

2 years ago

Which transformation could take place at the anode of an electrochemical cell?

erica [24]

As I found out the choices for your question which are:

<span>A) F2 to F-
B) Cr2O7²- → Cr2+
C) O2 to H2O
D) HAsO2 to As
</span>
Unfortunately, the answer does not belong to the choices provided. In fact, it is the oxidation half-reaction that occurs at the anode of an electrode for it to transform chemical energy to consumable electrical energy.

7 0

3 years ago

Define refractive index.​

Define refractive index.