<span>Following both the statements are true:
Some Defined it by
"Supported by most but not all".
Some defined it by this statement:
</span>"Not studied by many scientists"
Acceleration = (change in speed) / (time for the change)
Change in speed = (end speed) - (start speed)
Change in speed = (26 m/s) - (12 m/s) = 14 m/s
Time for the change = 6 s
Acceleration = (14 m/s) / (6 s)
Acceleration = (14/6) (m/s²)
<em>Acceleration = 2.33 m/s²</em>
Answer:
wen I was in the car toing home from school after a bad day n si si I have crazzyyy
Answer: -1 or -4/4
Explanation:
4/4 simplifies to 1 and the opposite of 1 is -1.
Answer:
3 AU
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 asteroid from the sun (orbital radius)
is the orbital period of the asteroid
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_a = \sqrt[3]{\frac{r_e^3}{T_e^2}T_a^2} =\sqrt[3]{\frac{(1 AU)^3}{(1 y)^2}(5.2 y)^2}=3 AU](https://tex.z-dn.net/?f=r_a%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_a%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281%20AU%29%5E3%7D%7B%281%20y%29%5E2%7D%285.2%20y%29%5E2%7D%3D3%20AU)
So, the distance of the asteroid from the Sun is exactly 3 times the distance between the Earth and the Sun.
