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uranmaximum [27]
3 years ago
11

Light travels at a constant speed of about 300,000 km/s. This speed is referred to as the speed of light. A light year is the di

stance that light travels in one year. One light year is equal to 9.46 trillion kilometers. The closest star to Earth, other than the Sun, is Proxima Centauri. It takes light from this star about 4.2 years to reach Earth. How far from Earth is the star Proxima Centauri? A. 4.2 light years B. 9.46 trillion km C. 300,000 light years D. 300,000 km.
Physics
2 answers:
gavmur [86]3 years ago
8 0

Answer:

The correct choice is

A. 4.2 light years

Explanation:

v = speed of light = 300,000 km/s

t = time taken for the light to travel from Proxima centauri to earth = 4.2 years

d = distance traveled by light in 1 year = 1 light year

So

distance in 1 year = 1 light year

distance in 4.2 year = 4.2 light year

Hence

distance of star Proxima Centauri from earth = 4.2 light year

Rama09 [41]3 years ago
7 0

Based on the information in the question, a light year is the distance traveled light in one year. Then if the light from the star Centauri takes 4.2 years to reach the earth, then its distance from earth is 4.2 light years.

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B

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An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0
3 years ago
If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, what
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Its total mechanical energy is <em>2,000 J</em>.

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Identify the prefix that would be used to express<br> 2,000,000,000 bytes of computer memory?
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4 0
3 years ago
Read 2 more answers
5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur
Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

   v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

   Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

  Л = 1 - \frac{Tl}{Th}  

where Tl = 300 k

            Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 )  :

= P3v3 = mRT3  ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

   = (1076.25 * 0.4 ) /  1237.2  

  = 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

        = 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below  

6 0
3 years ago
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