Is a-b parallel to c-d? Explain

A conservative estimate of the number of stars in the universe is 6 x 10^22. The average human can see about 3,000 stars at nigh

Ilia_Sergeevich [38]

$2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

<h3><u>Solution:</u></h3>

Given that, conservative estimate of the number of stars in the universe is $6 \times 10^{22}$

The average human can see about 3,000 stars at night with only their eyes

To find: Number of times more stars are there in the universe, compared to the stars a human can see

Let "x" be the number of times more stars are there in the universe, compared to the stars a human can see

Then from given statement,

$\text{Stars in universe} = x \times \text{ number of stars human can see}$

<em><u>Substituting given values we get,</u></em>

$6 \times 10^{22} = x \times 3000\\\\x = \frac{6 \times 10^{22}}{3000}\\\\x = \frac{6 \times 10^{22}}{3 \times 10^3}\\\\x = 2 \times 10^{22-3}\\\\x = 2 \times 10^{19}$

Thus $2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

5 0

4 years ago

Help please i need it fast

san4es73 [151]

Answer:

125x125

125=5x5x5

125x125=5x5x5x5x5x5=5 to the power of 6

6 0

3 years ago

Read 2 more answers

Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov

mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

$2t=(m+\frac{1}{2} ).\frac{\lambda}{n}$

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

$t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm$