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viva [34]
4 years ago
12

The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of po

sitions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Physics
1 answer:
creativ13 [48]4 years ago
3 0

Answer:

The distance is  d = 1.5 *10^{15} \ km

Explanation:

From the question we are told that

        The smallest shift is d = 0.2 \ grid \ units

Generally a grid unit is  \frac{1}{10} of  an arcsec

  This implies that  0.2 grid unit is  k =  \frac{0.2}{10} = 0.02  \  arc sec

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           d =  \frac{1}{k}

substituting values

           d =  \frac{1}{0.02}

           d = 50 \ parsec

Note  1 \ parsec  \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km

So  d = 50 * 3.08 *10^{13}

     d = 1.5 *10^{15} \ km

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La visualización molecular significa mirar modelos moleculares para explorarlos y comprenderlos. La visualización molecular no implica necesariamente el modelado molecular, lo que significa crear modelos moleculares o cambiar la composición o las configuraciones de los modelos existentes.
6 0
3 years ago
Help yea I need help
swat32
Answer is D I think....
8 0
3 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
3 years ago
uppose that the terminal speed of a particular sky diver is 150 km/h in the spread-eagle position and 320 km/h in the nosedive p
AysviL [449]

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\

\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55

4 0
3 years ago
Speed1=20 m/s. time1=5sec
lina2011 [118]

Answer:

<h2><em><u>2</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u>5</u></em><em><u> </u></em><em><u>m</u></em><em><u>/</u></em><em><u>s</u></em></h2>

Explanation:

<h2><em>AVERAGE</em><em> </em><em>S</em><em>P</em><em>E</em><em>E</em><em>D</em><em>=</em></h2>

<em>=  \frac{s1 t1+ s2t2 + s3t3}{t1 + t2 + t3}</em>

<em>=  \frac{20 \times 10 + 30 \times 5 + 15 \times 5}{10 + 5 + 5}</em>

<em>=  \frac{200 + 150 + 75}{20}</em>

<em>=  \frac{425}{20}</em>

<em>= 21.25 ms</em>

4 0
2 years ago
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