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Over [174]
3 years ago
11

6)the speed of light is approximately​ 186,000 mi/sec. It takes light from a particular star approximately 9 yrs to reach Earth.

How many miles away is the star from​ Earth? Express the answer in scientific notation. Use 365 days in 1 year. The star is nothing miles away from Earth.
Physics
1 answer:
NeTakaya3 years ago
5 0

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

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The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
3 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

\frac{x}{D}=\frac{1.22\lambda }{d}

where symbol's have their usual meaning

thus applying values we get

D=\frac{dx}{1.22\lambda }

\therefore D=\frac{0.633\times 4.61\times 10^{-3}}{1.22\times 547\times 10^{-9}}\\\\D=4372.77m\\=4.372km

5 0
3 years ago
A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block tha
Anna007 [38]

Answer:

v = 2.82 m/s

Explanation:

For this exercise we can use the conservation of energy relations.

We place our reference system at the point where block 1 of m₁ = 4 kg

starting point. With the spring compressed

        Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂

final point. When block 1 has descended y = - 0.400 m

        Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y

as there is no friction, the energy is conserved

       Em₀ = Em_f

       ½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y

       ½ k x² - m₁ g y = ½ m₂ v²

 

       v² = \frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y

let's calculate

        v² = \frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)

        v² = 2.7 + 5.23

        v = √7.927

        v = 2,815 m / s

using of significant figures

        v = 2.82 m/s

5 0
3 years ago
What negatively charged particle was discovered through the study of cathode rays?
tigry1 [53]
The particle discovered in cathode rays was the electron
5 0
3 years ago
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