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3 years ago

Help me with this please thank you

Physics

1 answer:

marshall27 [118]3 years ago

5 0

Birds have to go to south to migrate, birds are big and some are small, birds have good eye sit this is for the first one

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A source emits monochromatic light of wavelength 495 nm in air. when the light passes through a liquid, its wavelength reduces t

Katyanochek1 [597]

By definition, the refractive index is
n = c/v
where c = 3 x 10⁸ m/s, the speed of light in vacuum
v = the speed of light in the medium (the liquid).

The frequency of the light source is
f = (3 x 10⁸ m/s)/(495 x 10⁻⁹ m) = 6.0606 x 10¹⁴ Hz

Because the wavelength in the liquid is 434 nm = 434 x 10⁻⁹ m,
v = (6.0606 x 10¹⁴ 1/s)*(434 x 10⁻⁹ m) = 2.6303 x 10⁸ m/s

The refractive index is (3 x 10⁸)/(2.6303 x 10⁸) = 1.1406

Answer: a. 1.14

6 0

3 years ago

Hey please help Hey please help Hey please help Hey please help Hey please help

meriva

No they aren't the same

6 0

3 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

3 years ago

Describe a change caused by kinetic energy as well as a change that involves potential energy

kykrilka [37]

Kinetic energy is formed when the object is in motion.
Potential energy is the energy that is formed relative to others.

One of the example is Corn flour factory.

Corn turned into flour by a windmill that moved by the waterfall. Movement of the mill is relative to the power given by waterfall (potential energy) and the spinning crushes the corn into flour (kinetic energy)

5 0

3 years ago

Read 2 more answers

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.