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2 years ago

How do scientists determine motion of objects in the universe??

2 answers:

5 0

Usually, Scientists use "Parallax Method" in order to determine the different positions of an object

Hope this helps!

DerKrebs [107]2 years ago

3 0

Answer: By Doppler effect

Explanation:

The spectrum of the objects in the universe such as stars and galaxies is taken and studied. The apparent change in the frequency/wavelength of the light suggests the motion of these objects.

If the apparent frequency is greater, the light would be blue -shifted which means it is moving towards Earth and if it is red shifted, it would be moving away from the earth. This is apparent change in observed frequency is known as Doppler's effect which studied to determine the motion of objects in the universe.

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HELPPPPPPPPP Which option is a series of movement exercises for physical and mental health and is based on Hindu philosophy? O p

nika2105 [10]

Answer:

The correct answer is option ( D ) Yoga

Explanation:

hope it helps you dear :)

have a nice day

7 0

2 years ago

A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu

Ghella [55]

The angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

<h3>What is angular frequency?</h3>

Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.

<h3>What is cyclotron?</h3>

The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.

<h3>Calculation of angular frequency:</h3>

Given,

B = 0.47 T

r = 0.68

mass of proton = 1.6x $10^{-27}$

q = 1.6 x $10^{-19}$

so, the frequency is:

f = qB/2 $\pi$ m

f = 1.6 x $10^{-19}$ x 0.47/2x3.14x1.6x $10^{-27}$

f = 0.07 x $10^{8}$

Hence, the angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

Learn more about angular frequency here:

brainly.com/question/14244057

#SPJ4

7 0

1 year ago

In nuclear fission the nucleus is divided into two or more fragments, releasing Question 8 options: protons and energy. neutrons

Dmitrij [34]

Answer:

the answer is neutrons and energy

8 0

2 years ago

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

2 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

2 years ago