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spayn [35]
2 years ago
5

A coin is resting on the bottom of an empty container. The container is then fi lled to the brim three times, each time with a d

iff erent liquid. An observer (in air) is directly above the coin and looks down at it. With liquid A in the container, the apparent depth of the coin is 7 cm, with liquid B it is 6 cm, and with liquid C it is 5 cm. Rank the indices of refraction of the liquids in descending order (largest fist).
a. nC, nA, nB
b. nA, nB, nC
c. nB, nA, nC
d. nC, nB, nA
e. nA, nC, nB
Physics
1 answer:
ahrayia [7]2 years ago
3 0

when we placed in the container of real depth assuming = d

if the container filled with liquid A then its apparent depth d' = 7cm

so the refractive index nA = real depth / apparent depth

                                       = d/7cm = 0.1428d

if the container filled with liquid B then its apparent depth d' = 6cm

so the refractive index nB = real depth / apparent depth

                                       = d/6cm = 0.166d

if the container filled with liquid C then its apparent depth d' = 5cm

so the refractive index nC = real depth / apparent depth

                                       = d/5cm = 0.2d

Since the refractive index is inversely proportional to the apparent depth

then the refractive indices are nC > nB > nA

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Hope it helped you.

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4 0
3 years ago
If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?
Shalnov [3]

In a series circuit the total current is the same throughout resistors and so:

I_{total}=I_1=I_2

The voltage is distributed throughout the resistors and so:

V_{total}=V_1+V_2

and the total resistance can be calculated by adding up the resistors resistance:

R_{total}=R_1+R_2

First thing is to calculate the total resistance and so:

R_{total}=6\Omega + 4\Omega = 10\Omega

And by Omh's law V=IR we have:

V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.


6 0
3 years ago
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Who made advances to our understanding of the conscious and unconscious mind?
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4 0
3 years ago
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You move a 75-kg box 35 m. This requires a force of 90 N. how much work is done while moving the box?
Luda [366]
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4 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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