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3 years ago

What happens to a path of a light ray parallel to the principal axis, after it passes through a converging

1 answer:

3 0

Answer: The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. ... All three rays should intersect at exactly the same point.

Explanation: Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses.

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A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.

ruslelena [56]

The answer is 88m cause that’s the total distance the dog had ran in total and if that’s not the answer it’s 24m cause that’s how much further he had to go to retrieve it all the way back to the same position

3 0

3 years ago

An astronomer observes a hydrogen line in the spectrum of a star. The wavelength of hydrogen in the laboratory is 6.563 x 10-7m,

Sergio [31]

Answer:

D. The star is approaching Earth.

Explanation:

As we know by the doppler's effect of light that

$\frac{\Delta \nu}{\nu} = \frac{v}{c}$

here we know that

$\Delta \nu$ = change in frequency

here we know that the wavelength of light coming from the star is decreased so the frequency will increase

$\Delta \nu = \frac{c}{\lambda'} - \frac{c}{\lambda}$

$\Delta \nu =(3\times 10^8)(\frac{1}{6.56186 \times 10^{-7}} - \frac{1}{6.563 \times 10^{-7}})$

$\Delta \nu = 7.9414 \times 10^{10} Hz$

now we have

$\frac{7.9414 \times 10^{10}}{\nu} = \frac{v}{c}$

here we know that

$\nu = \frac{3\times 10^8}{6.563 \times 10^{-7}} = 4.57 \times 10^{14} Hz$

now we have

$v = 5.2 \times 10^4 m/s$

So here correct answer is

D. The star is approaching Earth.

8 0

4 years ago

What can a human still do better and faster than any machine learning solution<br>

emmasim [6.3K]

Answer:

mistakes I guess lol don't take serious

3 0

3 years ago

If a book has a a mass of 2 kg, how much does the book weigh?

Anestetic [448]

Answer:

A. 20 N

Explanation:

the weight of book is 2×10= 20N

7 0

3 years ago

A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative

Vesnalui [34]

If the current takes him downstream we must find the resultant vector of the velocities: $V res= \sqrt{1^{2}+0.91^{2} } = \sqrt{1.8281}= 1.3520747$ Then if the river is 3000 m-wide the swimmer will have to pass:
1.3520747 · 300 = 4056.14 m t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.

5 0

3 years ago