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3 years ago

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un objeto de 10kg cae desde una altura y tarda 2 segundos en caer al suelo. cual era su velocidad al llegar al suelo? de que alt

ura cayo?

1 answer:

const2013 [10]3 years ago

8 0

Explanation:

hru3he. djehe isbs she house siege dbsus d beuh

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what were the negative results with the american television from analog broadcast to digital broadcast?

alexandr1967 [171]

Answer:

Explanation:

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8 0

3 years ago

Read 2 more answers

Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the

True [87]

Answer:

$E=54V/cm$

$\Delta V=496.8V$ between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

$\Delta V=Ed$

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

$E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm$

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

$\Delta V=Ed=(54V/cm)(9.2cm)=496.8V$

8 0

3 years ago

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is $4.8 \cdot 10^{-6}\,mg$

Explanation:

The ration between Uranium mass and total sample mass is: $\frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}$

For a sample of mass 1.2 mg, the amount of uranium is:

$1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg$

7 0

3 years ago

When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at

Veseljchak [2.6K]

Answer:

First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c). The value of the refractive index of the medium is a measure of its "optical density": Light spreads at maximum speed in a vacuum but slower in others transparent media; therefore in all of them n> 1. Examples of typical values of are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).

The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:

The limit or minimum angle, α lim, is defined as the angle of refraction from which the refracted ray disappears and all the light is reflected. As in the maximum value of angle of refraction, from which everything is reflected, is βmax = 90º, we can know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:

βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1

Explanation:

When a light ray strikes the separation surface between two media different, the incident beam is divided into three: the most intense penetrates the second half forming the refracted ray, another is reflected on the surface and the third is breaks down into numerous weak beams emerging from the point of incidence in all directions, forming a set of stray light beams.

4 0

3 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

3 years ago