Answer:
Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?
Liquid is the answer
Explanation:
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Part g)

Explanation:
Initial speed of the launch is given as
initial speed = 
angle =
degree
Now the two components of the velocity

similarly we have

Part a)
Now we know that horizontal range is given as

maximum height is given as

so we have

time of flight is given as



Part b)
Now the speed of the ball in x direction is always constant
so at the peak of its path the speed of the ball is given as



Part c)
Initial vertical velocity is given as


Part d)
Initial speed is given as

so we will have


Part e)
Angle of projection is given as



Part f)
If we throw at same speed so that it reach maximum height
then the height will be given as


Part g)
For maximum range the angle should be 45 degree
so maximum range is


Water expands when it freezes (that's why you should never put closed, fully filled water bottles in the freezer !)
I believe flowing water changes the land because eventually frozen water has to melt to flowing water, right?
Bernini's sculpture "Apollo and Daphne" implies a chase scene motion. Apollo and Daphne<span> is a life-sized Baroque marble </span>sculpture<span> by Italian artist Gian Lorenzo </span>Bernini<span>, executed between 1622 and 1625. Hope this answers the question. Have a nice day.</span>