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Anna11 [10]
3 years ago
11

Anyone know how to do this?

Physics
1 answer:
MrMuchimi3 years ago
7 0

The voltage from one side of the battery all the way around to the other side of the battery is 12v .

If 4 of those volts show up across the circle-thing, then the rest of the 12v ... 8v ... Must show up across the set of parallel rectangles.

To get that answer, I subtracted the 4 from the 12.

Just like it says in choice-C.

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A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf
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Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\

where <em>nw</em> and <em>na</em> is the refractive indices of water (1.33) and air (1.00) respectively; <em>s</em> is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0

Now we can solve for the apparent depth of the fish.

(a) s'=-(\frac{n_{a}}{n_{w}})x s (Make s' subject of the formula from the above equation)

s'=(\frac{1.00}{1.33} )x7cm

∴ s'=5.3 cm.

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

s'=-(\frac{n_{a}}{n_{w}})x s

s'=(\frac{1.00}{1.33} )x 33 cm

s'=24.8 cm.

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

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