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Aliun [14]
2 years ago
15

20" id="TexFormula1" title="\huge\bold{\purple{\bold{⚡Gravitational Constant?⚡}}} " alt="\huge\bold{\purple{\bold{⚡Gravitational Constant?⚡}}} " align="absmiddle" class="latex-formula"> ​
Physics
2 answers:
baherus [9]2 years ago
8 0

\huge\underline\mathtt\colorbox{cyan}{G=}

6.673 \times  {10}^{ - 11}

And unit is Nm^2/kg^2

crimeas [40]2 years ago
6 0

Explanation:

The attractive force between two bodies when multiplied by the product of the masses of two bodies and divided by the square of the distance between them.

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Answer:

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Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

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\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

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\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

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