Mercury has a density of 13.6 g/cm , whereas the density of water is 1.00 g/cm. The pressure of the atmosphere is high enough to push mercury 76.0 cm up
through a vacuum tube. How high can a column of water be pushed up in a vacuum tube by the atmosphere? (Hint: The level of either liquid rises till the atmospheric
pressure is in balance with the pressure by the liquid computed by density xgx height; since the atmospheric pressure is the same in both scenarios, one only
needs to compare the liquid pressures to find the height of water) (Express the height of water in meters and with 3 significant figures in the form XX. X; number only
- no units)

10.3

Explanation:

Step 1:

The pressure exerted by any liquid column of height, h density d is given by the formula P = h * d * g

Step 2:

It is given that one atmosphere pressure pushes up 76.0 cm of mercury, we need to calculate the level of water that will be pushed by the same pressure.

Step 3:

Since the pressure pushing up mercury and water is the same

* * g = * * g

  =  = (13.6 g/cm * 76 cm)/1 g/cm = 1033.6 cm

Step 4:

Now we need to express the answer in meters.

1 m = 100 cm.

1033.6 cm = 10.336 m

This can be rounded off to 10.3 m

the answer would be   dignified - apex

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .