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3 years ago

What is an example of total internal reflection at work?

1 answer:

6 0

Total internal reflection occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface. <span>Total internal reflection can be observed while swimming, when one opens one's eyes just under the water's surface. If the water is calm, its surface appears mirror-like.</span>

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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

4 0

4 years ago

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.)