An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

where;

height h which is the height for the free surface in a rotating tank is expressed as:

at the bottom surface of the tank;

r = 0, h = 0

∴

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

replacing the value of h in equation (2); we have:

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then

Replacing equation (1) and (3)

Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

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 .

a sonic boom is a loud boom that breaks the sound barrier

explanation: