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Dovator [93]
3 years ago
14

a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus

t show false colors? g
Physics
1 answer:
aivan3 [116]3 years ago
6 0

Answer:

The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.

Explanation:

  • Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.
  • As, these telescopes includes the most recent "Hubble telescope", which is out there inside the space to explore the universe and more over the galaxies by subjecting them with x-rays and then provide us with a very rough but valid results to study the distant galaxies.

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"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro
pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0
3 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
Joe Burrow of the Cincinnati Bengals ate a turkey sandwich and an apple for lunch. Later that day, he spent 2 hours on the footb
Nostrana [21]

Answer:

A Thermal energy was converted to kinetic energy

7 0
2 years ago
The net force acting on an object equals the applied force plus the force of friction.
Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0
2 years ago
A compact disc (CD) is played by a cd player, which uses a laser to read the tracks on the disc. The disc spins initially at app
uranmaximum [27]

Answer:

a. The laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Explanation:

a. The laser tracking mechanism experiences a changing tangential velocity

This is because the tangential velocity v = rω where r = radius of disc and ω  = angular speed of discs. Since r is constant, v ∝ ω.

Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.

So, the laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.

Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.

So, The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.

Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.

So, the laser tracking mechanism experiences a non-zero tangential acceleration

Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.

4 0
2 years ago
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