Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
I believe they are:
Pin Insulators
Suspension Insulators
Strain Insulators
If thats not enough theres also insulators commonly used of lower voltage which are:
Stay
Shackle
I believe the correct answer from the choices listed above is the last option. If the volatility of X is higher than that of Y, then <span>Y’s molecules experience stronger London dispersion forces than X’s molecules. All molecules has london dispersion forces. Also, the stronger the bond, the harder it is to volatilize. Hope this answers the question.</span>
Since they are made of different chemical make ups, boiling point and density they will most likely be in different groups on the periodic table.
