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Jlenok [28]
4 years ago
15

Earth orbits 1 AU from the Sun, and the Oort cloud extends from about 10,000 to 100,000 AU from the Sun. If you represent Earth’

s orbit around the Sun with a paper plate 4 inches in radius, how far away will the inner edge of the Oort Cloud be? The outer edge? Express your answers in miles.
Physics
1 answer:
sertanlavr [38]4 years ago
4 0

Explanation:

Earth’s orbit around the Sun represented as :

1 AU = 4 inches

Therefore,

<u>Inner edge</u> of the Oort Cloud represented as :

10,000 AU = 40,000 / 63360 = 0.631 miles

<u>Outer edge</u> of the Oort Cloud represented as<u>:</u>

100,000 AU = 400,000 / 63360 = 6.31 miles

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How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase
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Answer:

The value is  v_o =  5.488 \  m/s

Explanation:

From the question we are told that

     The emitted frequency increased by \Delta f =  1.6 \% = 0.016 \

Let assume that the initial value of the emitted frequency is

      f =   100\%  =  1

Hence new frequency will be  f_n  =  1 +  0.016 = 1.016

Generally from Doppler shift equation we have that

         f_1 =  [\frac{ v \pm v_o}{v \pm + v_s } ] f

Here v  is the speed of sound with value  v =  343 \ m/s

         v_s is the velocity of the sound source which is v_s = 0 \ m/s because it started from rest

         v_o  is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

                   f_1 =  [\frac{ v + v_o}{v + 0 } ] f

=>                1.016  =  [\frac{ 343  + v_o}{343 } ] * 1  

=>                v_o =  5.488 \  m/s

         

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Cual es la densidad de una sustancia si pesa 24,69 gr y ocupa un volumen de 33,1 centímetros cúbicos si la formula de densidad e
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Answer:

In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.

Explanation:

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TEA [102]

Answer:

Option (B)

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