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3 years ago

What stage do most stars spend the majority of their lives in?

2 answers:

8 0

It a stage known as the "Main Sequence" stage. Stars achieve nuclear stability at this point and remain in this stage until the end of their lives.
Have a nice day!

hjlf3 years ago

3 0

Out of the 7 most important stages a star spends about 90% of it's life in the "Main Sequence Star" stage. In this stage a star forms helium in it's core by fusing hydrogen atoms. Actually the Sun is in this stage right now. How long it stays in this stage depends on the star's size.

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According to the periodic table, the average atomic mass of helium is A) 2 amu. B) 2.0026 amu. C) 4.0026 amu. D) 6.0026 amu.

RUDIKE [14]

Answer:

C) 4.0026 amu

Explanation:

Helium (He) is the 2nd element on the Periodic Table.

It's neutral atom has 2 protons and 2 electrons.

It is in the 1st period and the 18th row.

It is also a Noble gas.

3 0

3 years ago

A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has

Masja [62]

Answer:

$v_f=7*10^7\frac{m}{s}$

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

$v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}$

5 0

4 years ago

What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum fo

Pachacha [2.7K]

Answer:

14,300 lines per cm

Explanation:

Answer:

14,300 cm per line

Explanation:

λ400 nm to 400nm

We can find the maximum number of lines per centimeter, which is reciprocal of the least distance separating two adjacent slits, using the following equation.

mλ = dsin (θ)

In this equation,

m is the order of diffraction.

λ is the wavelength of the incident light.

d is the distance separating the centers of the two slits.

θ is the angle at which the mth order would diffract.

To find the least separation that allows the observation of one complete order of spectrum of the visible region, we use the maximum wavelength of the visible region is 700 nm.

d = mλ / sin (θ)

As we want the distance d to be the smallest then sin (θ) must be the greatest, and the greatest value of the sin (θ) is 1. For that we also use the longest wavelength because using the smallest wavelength, the longest wavelength would not be diffracted.

d = mλ / sin (θ)

d = 1 x 700nm / 1

= 700 nm

So, the least separation that would allow for the possibility of observing complete first order of the visible region spectra is 700 nm, and knowing the least separation we can find the maximum number of lines per cm, which is the reciprocal of the number of lines per cm.

n = 1/d

= 1 / 700 x $10^{-9}$