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2 years ago

2. The light from the Sun reaches Earth in 8.3 minutes. The speed of light is 11,184,681 miles/minute. How far is

Physics

1 answer:

12345 [234]2 years ago

6 0

Answer:

92,832,852.3 miles

Explanation:

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$

hence, the displacement of the airplane is 3.45*10^4 m

6 0

3 years ago

The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

3 years ago

We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to

Zigmanuir [339]

Answer:

gamma rays , X rays, ultraviolet , visible light , infrared, radio waves

Explanation:

The electromagnetic spectrum is the set of electromagnetic radiations distributed in their different frequencies or wavelengths, which in turn are related to their energy. If we go from the smallest wavelengths known up to now (because according to physics the electromagnetic spectrum is infinite and continuous) to the longest, the electromagnetic spectrum covers the following radiations:

Gamma rays, X-rays, ultraviolet, visible light (all the colors we are able to see), infrared, radio waves and microwaves.

Let's make a brief of them:

-Gamma rays: With a wavelength in the order of $10^{-12}m$ , is a type of ionizing radiation capable of penetrating matter quite deeply and is able to cause serious damage to the nucleus of the cells. Inaddiito, these rays are used to sterilize medical equipment and food.

-X rays: With a wavelength between $1m$ and $10km$ . It is invisible to the human eye, capable of crossing opaque bodies and of being an ionizing radiation.

-Ultraviolet light: Whose wavelength is approximately between 100 nm and 380 nm; is a type of electromagnetic radiation that is not visible to the human eye.

-Visible light: This part of the spectrum is located between ultraviolet light and infrared light (380 nm - 780 nm). It should be noted, the fact the only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

-Infrared: This type of radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm).

These waves can be divided into:

- Near infrared or long wave infrared: it is the least sensitive to color and is easily absorbed by water.

- Medium or medium wave infrared: it is also insensitive to color and easily absorbed by water and many types of plastics and paints.

- Far infrared or short wave infrared: it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.

This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

-Radio waves: These are a type of electromagnetic radiation with wavelengths between 10 m to 10,000 m. This type of electromagnetic waves is very well reflected in the ionosphere, the layer of the atmosphere through which they travel directly or using repeaters. In addition, they are very useful to transport information, being important in telecommunications. They are used not only for conventional radio transmissions but also in mobile telephony and TV.

5 0

3 years ago

2. An alternating current is represented by the equation I=20sin 100mt.

Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

$I=I_o sin\omega t$