An object of mass 2.00 kg is held at a position a, a vertical height of 20.0 m above the ground. point b is 8.00 m directly below a. neglect air resistance and use g = 10.0 m/s2. what is the speed of the object at position b when it falls from position a? i have no idea how to go about

find the time for the object to fall from position A to position B. You can use \[x = \frac{1}{2}at^2\]to do so. \(x\) represents the distance traveled, \(a\) is the acceleration (here due to gravity, so use the value provided for \(g\)), and \(t\) is the time. When you have found the time required for the object to fall from position A to position B, you can use \[v = at\] to calculate the speed. Again, \(a=g\) and \(t\) is the value you found in the first part ;)

Idon't know it i just need my points lol

The answer is water molecules are split.cellular respiration and photosynthesis are complement  reactions to each other. these are reverse reactions. in cellular respiration glucose and oxygen yield water and carbon dioxide. so, when respiration begins,  glucose molecules are split. in the opposite reaction - in photosynthesis, carbon dioxide and water yield glucose and oxygen. thus,  when respiration begins, water molecules are split.