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drek231 [11]
3 years ago
7

1. Have you ever tried to undergo an X- ray to test on youur body? What can you see in the fiom after the examination? Why is th

e bone being visible and not organs? 2.How do airports and some transportation stations detect a person affecteed by a virus? 3. What happen to white light as it hits a prism? plsss answer my question po0 pllssssssssss................
Physics
1 answer:
liraira [26]3 years ago
5 0

Answer:

1) λ < 2d,  2)  nfrared imaging technique, 3) each color there is a different index of refraction

Explanation:

We are going to answer the three questions

1) When x-rays pass through matter in order to be dispersed, their wavelength must be of the order of the length of separation in the atoms and molecules of the body, in solid bones this length is similar and they scatter and reflect the x-rays therefore they can be observed, the fat and the soft tissue have a much greater separation therefore the x-rays cannot be reflected and consequently it is not observable by this technique.

2) At airports they use the infrared imaging technique, where the image is taken for the infrared wavelength, which is the heat part of the electromagnetic spectrum; consequently, when the image is viewed, the hottest areas appear brighter and, since when a person has a virus, his temperature rises, his temperature rises, it is possible to observe people with a higher temperature.

3) when white light hits a prism it is refracted with the equation

            n₁ sin θ₁ = n₂ sin θ₂

where the incidence of refraction depends on the wavelength, therefore for each color there is a different index of refraction and consequently the light is separated in its different colors.

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Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou
spin [16.1K]

Answer:

(a) 28690 m/s (b) 2.46x10^{33}J

Explanation:

The orbital speed is define as:

v = \frac{2 \pi r}{T}   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

v = \frac{d}{t}   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

d = vt   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

8min . \frac{60s}{1min} ⇒  480s

t = 480s

Replacing all those values in equation 3 it is gotten:

d = (3.0x10^{8}m/s)(480s)

d = 1.44x10^{11}m

Kepler’s third law is defined as:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

T = \sqrt{r^{3}}

T = \sqrt{(1.44x10^{11}m)^{3}}

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

T = \sqrt{1AU}

T = 1AU

That can be expressed in units of years.

T = 1AU . \frac{1year}{1AU}

T = 1year

But there are 31536000 seconds in one year:

T = 1year . \frac{31536000s}{1year}

T = 31536000s

Finally, equation  1 can be used:

v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}

v = 28690 m/s

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

E = \frac{1}{2}mv^{2}  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

F = \frac{GMm}{r^{2}}

ma = \frac{GMm}{r^{2}}  (5)

M will be isolated in equation 5:

M = \frac{r^{2}a}{G}

Where r is the radius of the Earth (6.38x10^{6}m)

M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})

M = 5.98x10^{24} Kg

E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}

E = 2.46x10^{33}Kg.m^{2}/s^{2}

E = 2.46x10^{33}J

<u>Hence, the kinetic energy of Earth is 2.46x10^{33}J.</u>

8 0
3 years ago
A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65
salantis [7]

One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

\epsilon = NBA\omega

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

\omega = Angular velocity

Re-arrange to find N,

N = \frac{\epsilon}{BA\omega}

Our values are given as,

\epsilon = 19V

B = 0.434T

\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s

A = (6.65*10^{-2})^2 m^2

Replacing at our equation we have:

N = \frac{\epsilon}{( 0.434)A\omega}

N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}

N = 31.63 \approx 32

Therefore the number of loops of wire should be wound on the square armature is 32 loops

6 0
3 years ago
What do the movements of stars and galaxies tell astronomers about how the universe formed?
natulia [17]
This could be Hubble's law, or something related to it. I think there's a possibly Doppler RED SHIFT in the optical spectra of stars etc as observed on the earth. It seems that they are accelerating away from the earth, and that the further away they are the faster they are moving.
It seems that this has been connected to the idea of "The Big Bang" theory of the origin of the universe which seems to have superceded Professor Sir Fred Hoye's Steady State theory of the universe.
There's some Special Relativity in this lot, too.
3 0
3 years ago
What linear speed must an earth satellite have to be in a circular orbit at an altitude of 159 km?
IgorLugansk [536]
Gravitational acceleration, g = GM/r^2. Additionally, for a satellite in a circular orbit, g = v^2/r

Where, G = Gravitational constant, M = Mass of earth, r = distance from the center of the earth to the satellite, v = linear speed of the satellite.

Equating the two expressions;
v^2/r = GM/r^2
v = Sqrt (GM/r);
But GM = Constant = 398600.5 km^3/sec^2
r = Altitude+Radius of the earth = 159+6371 = 6530 km

Substituting;
v = Sqrt (398600.5/6530) = 7.81 km/sec = 781 m/s
8 0
3 years ago
The critical angle for water is 49°. If a ray of light
Sonja [21]

Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49      Ni = 1.325 deg     the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg  

So refraction can occur  outside the pond and you do not have total internal refection.

 

3 0
3 years ago
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