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Ipatiy [6.2K]
3 years ago
5

If light travels from oil (slower medium) to water (faster medium) at an angle, what happens to the direction of the light ray i

n water with respect to normal?
A. It moves away from the normal. B. It moves toward the normal.
C. It will move along the normal. D. It will move perpendicular to the normal.
E. It stops traveling.
Physics
1 answer:
iVinArrow [24]3 years ago
7 0
If light travels from oil to water at an angle, what happens to the direction of the light ray in water with respect to the normal, is it moves away from the normal.
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Electrons

Explanation:

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A runner achieves a velocity of 11.1 m/s, 9 sec after he begins. What is his
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Answer:

1.23 m/s²

Explanation:

Given:

v₀ = 0 m/s

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The runner's acceleration is 1.23 m/s².

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You take a couple of capacitors and connect them in series, to which you observe a total capacitance
Zepler [3.9K]

Answer:

Approximately \rm 5.7\; \mu F and approximately 29\; \rm \mu F.

Explanation:

Let C_1 and C_2 denote the capacitance of these two capacitors.

When these two capacitors are connected in parallel, the combined capacitance will be the sum of C_1 and C_2. (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)

C(\text{parallel}) = C_1 + C_2.

On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:

\displaystyle \frac{1}{C(\text{series})} = \frac{1}{C_1} + \frac{1}{C_2}.

(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field (V) between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored (Q) stays the same.)

The question states that:

  • C(\text{parallel}) = 35\; \rm \mu F, and
  • C(\text{series}) = 4.8\; \rm \mu F.

Let the capacitance of these two capacitors be x\; \rm \mu F and y\; \rm \mu F. The two equations will become:

\displaystyle \left\lbrace \begin{aligned}& x + y = 35 \\ & \frac{1}{x} + \frac{1}{y} = \frac{1}{4.8}\end{aligned}\right..

From the first equation:

y = 35 - x.

Hence, the y in the second equation here can be replaced with (35 - x). That equation would then become:

\displaystyle \frac{1}{x} + \frac{1}{35 - x} = \frac{1}{4.8}.

Solve for x:

\displaystyle \frac{x + (35 - x)}{x \, (35 - x)} = \frac{1}{4.8}.

x\, (35 - x) = 4.8.

x^2 - 35 \, x + 168 = 0.

Solve this quadratic equation for x:

x \approx 5.7 or x \approx 29.3.

Substitute back into the equation y = 35 - x for y:

  • x \approx 5.7 and y \approx 29.3, or
  • x \approx 29.3 and x \approx 5.7.

In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately 5.7\; \rm \mu F and approximately 29\; \rm \mu F (both values are rounded to two significant digits.)

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