<h2>T
he small mass should be placed at x = 11.72 cm or x = 68.28 cm </h2>
Step-by-step explanation:
Let the small mass be m and position on x axis be y.
A 20 kg sphere is at the origin.
Distance to 20 kg mass = y
A 10 kg sphere is at x = 20 cm
Distance to 10 kg sphere = 20 - y
We have forces between them are equal
We have gravitational force
Where G = 6.67 x 10⁻¹¹ N m²/kg²
M = Mass of body 1
M = Mass of body 2
r = Distance between them
Here we have
![\frac{G\times 20\times m}{y^2}=\frac{G\times 10\times m}{(20-y)^2}\\\\800-80y+2y^2=y^2\\\\y^2-80y+800=0\\\\y=68.28cm\texttt{ or }y=11.72cm](https://tex.z-dn.net/?f=%5Cfrac%7BG%5Ctimes%2020%5Ctimes%20m%7D%7By%5E2%7D%3D%5Cfrac%7BG%5Ctimes%2010%5Ctimes%20m%7D%7B%2820-y%29%5E2%7D%5C%5C%5C%5C800-80y%2B2y%5E2%3Dy%5E2%5C%5C%5C%5Cy%5E2-80y%2B800%3D0%5C%5C%5C%5Cy%3D68.28cm%5Ctexttt%7B%20or%20%7Dy%3D11.72cm)
So the small mass should be placed at x = 11.72 cm or x = 68.28 cm
Answer:
<em>1</em><em>8</em><em>0</em><em>-</em><em>(</em><em>6</em><em>3</em><em>+</em><em>6</em><em>3</em><em>)</em>
<em>=</em><em>1</em><em>8</em><em>0</em><em>-</em><em>1</em><em>2</em><em>6</em>
<em>=</em><em>5</em><em>4</em>
<em>1</em><em>8</em><em>0</em><em>-</em><em>(</em><em>3</em><em>7</em><em>+</em><em>9</em><em>0</em><em>)</em>
<em>1</em><em>8</em><em>0</em><em>-</em><em>1</em><em>2</em><em>7</em>
<em>5</em><em>3</em>
<em>1</em><em>8</em><em>0</em><em>-</em><em>(</em><em>5</em><em>4</em><em>+</em><em>5</em><em>3</em><em>)</em>
<em>=</em><em>1</em><em>8</em><em>0</em><em>-</em><em>1</em><em>0</em><em>7</em>
<em>=</em><em>7</em><em>3</em><em>°</em>
the answer to that question is the first choice
Answer:
G-$3000, L- $1200, M- $1800
Step-by-step explanation:
Georgia= 5/10 * 6000= $3000
Leanee= 2/10 * 6000= $1200
Maya= 3/10 * 6000= $1800