Answer:
How much time does his victim on the ground below have to move out of harm's way? At what velocity will the safe hit the ground? sownt d-200m. n a = 100/2. Vi-o.
Explanation:
There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"
The art event called “cant help myself” its very emotional if you understand it :)
ANOTHER RUNNING DOG
Explanation:
In the given question it is to find a suitable reference point to describe the motion of dog. Here I could suggest that it is better to compare the dog with another running dog to create the relative speed difference to get a reliable motion variation.
Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to another dog which is already in motion.
Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with another dog running.
The period of the planet is 58.1 years
Explanation:
We can solve this problem by using Kepler's third law, which states that:
"The square of the orbital period of a planet is proportional to the cube of its semimajor orbital axis"
Translated into equations, we can write:
![\frac{T_x^2}{r_x^3}=\frac{T_e^2}{r_e^3}](https://tex.z-dn.net/?f=%5Cfrac%7BT_x%5E2%7D%7Br_x%5E3%7D%3D%5Cfrac%7BT_e%5E2%7D%7Br_e%5E3%7D)
Where, taking the orbits of the planets as almost circular, we have:
is the orbital period of the unknown planet
is the radius of the orbit of the planet
(1 year) is the period of the orbit of the Earth
is the orbital radius of the Earth
Solving for
, we find the orbital period of the unknown planet:
![T_x = T_e \sqrt{\frac{r_x^3}{r_e^3}}=(1y)\sqrt{\frac{(15.0)^3}{(1.0)^3}}=58.1 y](https://tex.z-dn.net/?f=T_x%20%3D%20T_e%20%5Csqrt%7B%5Cfrac%7Br_x%5E3%7D%7Br_e%5E3%7D%7D%3D%281y%29%5Csqrt%7B%5Cfrac%7B%2815.0%29%5E3%7D%7B%281.0%29%5E3%7D%7D%3D58.1%20y)
Learn more about Kepler's third law:
brainly.com/question/11168300
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