We need Bernoulli's equation to solve this problem:

The problem statement doesn't tell us that the height changes, so we can remove the last term on each side of the expression, then arrange to solve for the final pressure:

We know the initial pressure, so we still need to calculate the initial and final velocities. We'll use the continuity equation:

Rearrange for velocity:

Where  is the cross-sectional area. We can calculate this for each diameter of the tube:

Now we can calculate the velocity for each diameter:

Now we have all of the values needed for Bernoulli's equation, allowing us to solve:

To begin with, it will be necessary to make use of Bernoulli's equation:

For the situation described in the question stem, we'll designate the top of the container as point 1, and the hole where water is flowing out as point 2.

To begin simplifying things, it's important to realize a few things. First, both points are open to the atmosphere. Therefore, the  term on each side of the above equation is equal to 1atm and thus can cancel out. Secondly, since the size of the hole on the side of the tank is so small compared to the rest of the tank, the velocity of the water at point 1 is nearly equal to 0. Hence, we can cancel out the  term on the left side of the equation. Thus far, we have:

Dividing everything by , we obtain:

And rearranging:

The equation relating fluid pressure, kinetic energy, and potential energy from state to state is known as the Bernoulli equation, and is as follows:

Our potential energy is the same, so we can remove that part from the equation.

 is the density, which in our case is equal to 1, so it doesn't change anything here.

We have the values for the initial pressure, initial velocity, and final velocity, so we can rearrange our equation to equal final pressure.

Now, we can plug in our values.

Therefore, the final pressure is equal to .

This is solved using Bernoulli's equation and the definition of pressure. First choose the "Bernoulli points", one just inside the roof where the air is still (Point A) and one just outside where the air is moving (Point B). This will allow us to eliminate many of the terms:

Since the air is still inside, 

Since our points are at the same height, the  terms cancel. Rearrange:

Since we only care about the difference in pressure inside and out.

Plug in known values and solve for the difference in pressure.

Force is pressure times area:

Use Bernoulli's equation to find the pressure difference on the two sides of the kite. Call point A the inner surface (where the air is still) and point B the outer surface (where the wind is at full speed).

Since the points are at the same height, the  terms cancel. Since the air is still at point A, the  term is zero. Since we only care about the difference in pressure on the two sides of the kite, solve for :

Net force is pressure difference times area:

The moving water on one side of the metal plate has a lower pressure than the still water on the other side, resulting in a force.  We start by writing the Bernoulli equation:

We choose our two "Bernoulli points" to make the problem as easy as possible. Take point A to be on the side where the water is still, and point B on the side where the water is moving. If we make them at the same height, the  terms can be subtracted from both sides. Since the water is still at point A, the velocity term on the right-hand side of the equation is zero. Rearrange to find the difference in pressure from side A to side B:

This is saying that the pressure at point B is  less than the pressure at point A. Using the definition of pressure, find the resulting force:

We will use Bernoulli's equation to solve this. We must do this twice: once for the air and once for the water. The central principle here is that the moving stream of air has a lower pressure than still air. In this problem we will ignore the atmospheric pressure since it is applied at the tube ends and at the surface of the water outside the vertical tube.  For the air, choose our two "Bernoulli points": point A is just outside the horizontal tube and point B is just inside. Write the equation:

The heights are the same, so they cancel out of the equation. The air is still at point A, so the velocity term is zero for the left side. Finally, as mentioned, we care only about the difference in pressure:

This is saying that the pressure inside the tube is  below the pressure outside.

Now we solve for the water. Our points will be on the surface outside the vertical tube (point A) where the pressure is one atmosphere, and inside the vertical tube at the surface of the risen column (point B). Write the equation:

Since the water is no longer moving, the terms containing  are equal to zero. Rearrange:

Put in numbers and solve for the height difference:

Since Bernoulli's principle is derived from the work-energy theorem, it is a requirement that the flow be incompressible, steady, and without internal friction. Otherwise, energy would be lost to these outlets and the equation would no longer be applicable. 

In the question stem, we're told that a fluid with a density of  moves through a pipe at a speed of  and has a pressure of . We're then told that the same fluid begins to move through the pipe at a new speed of , and we're asked to determine what the new pressure will be.

In order to answer this question, we'll need to make use of Bernoulli's equation. This equation essentially tells us that the pressure, kinetic energy, and potential energy of a moving fluid is constant. Or, put another way:

Alternatively, since we know the sum of these values is constant, we can relate the sum of these values at one instant to the sum of these values at another instant.

Furthermore, since we're told that the fluid remains flowing in a horizontal pipe, the height of the fluid does not change. Therefore, we can cancel out the potential energy terms on both sides.

Next, if we define the initial conditions as instant 1, and the final conditions as instant 2, then the term we are trying to solve for is .

Then, plugging in the values we have allows us to obtain the answer:

To solve this problem, we will use Bernoulli's equation, a simplified form of the law of conservation of energy. It applies to fluids that are incompressible (constant density) and non-viscous.

Bernoulli's equation is: 

Where  is pressure,  is density,  is the gravitational constant,  is velocity, and  is the height. 

In our question, state 1 is at the tap and state 2 is at the nozzle. Input the variables from the question into Bernoulli's equation: