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)
Answer:
vpg = 0.064 N
Explanation:
Upthrust = Volume of fluid displaced
upthrust liquid on the cube g=10ms−2
vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N
vpg = 0.064 N
hope it helps.
I believe the answer is D. Thunderstorm
Unstable air indicates that <span>the lowest layers of an </span>air<span> mass is low in temperature. Thunderstorms is more likely to happen if there is unstable air mass because it provide a situation where there is more energy for the Thunderstorms to feed off of.</span>
