Pressure - AP Physics 2

We can begin this problem by writing the equation for pressure.

We have to remember that the value of atmospheric pressure is equal to 1atm, which is also equal to 101325Pa.

Now, if we set the gauge pressure equal to three times the atmospheric pressure, we can solve for the depth.

Pressure increases with depth according to:

Since the pressure at the surface of the ocean is the same as the pressure inside the bathysphere, we can ignore both. (This is called gauge pressure, or pressure above one atmosphere). Find the pressure at our depth:

Find the area of the hatch, and use the definition that force is pressure times area:

Pressure in a static fluid varies along the axis in which gravity acts. In other words, the deeper one goes, the higher the pressure is. This can be expressed as:

, where h represents the depth. Since we are examining a point 0.1m from the bottom, this is a depth of 0.9m; 0.5 of which is through oil and 0.4 of which is through water.

The pressure in a tire is a gauge pressure; the 25psi is the difference between the pressure in the tire and atmospheric pressure, and does not add atmospheric pressure to reach the measured number. Absolute pressure adds these quantities together.

The interface between the liquids occurs at the bottom of the tube. Here, the pressure is the same. We also know that the tube is open to atmospheric pressure at the top of each side of the tube. Thus, we can create two equations for the pressure at the interface and set them equal to each other:


By setting these equal to one another the result is:

Since the density of water is greater than that of oil, we can conclude that the height of the oil column will be greater than that of the water.

The pressure at the lake's surface is just the atmospheric pressure, . The pressure at depth below the surface is . Therefore, the pressure difference is given as .

We can then substitute the values in to arrive at the answer in :

Recall that this pressure is known as gauge pressure, and that the total pressure at the point underwater in the lake includes gauge pressure and atmospheric pressure.

This question is presenting us with a situation in which an object (a person) submerged in a liquid is experiencing pressure. To solve for the correct pressure, we'll need to take a few things into account.

First, we'll need to consider the density of the liquid. In this case, we're told that the liquid is saltwater, and we're given the density in the question stem. Second, we'll need to consider how far below the surface of the liquid our object is. Again, this value is given to us in the question stem. And third, we also need to remember that there is a pressure above the liquid, which is the pressure coming from the atmosphere. This value (which should be committed to memory) is , or .

Putting all these considerations together, we have an equation that we can use to calculate the pressure.

Plug in the values that we know and solve for total pressure.

This is a fairly straight-forward question, providing us with a force vector that is acting upon a surface and asking us for the pressure.

First and foremost, we can examine the equation for pressure:

Next, we'll need to figure out the sign convention for pressure. Based on the above equation, it may seem like if we plug in a negative value for pressure and a positive value for area, we'll get a negative value for pressure. However, this is not the case.

As it turns out, pressure is a scalar quantity. The reason for this is that when any force acts on a surface, it is acting perpendicularly to that surface. Therefore, the direction in which the force is acting is dependent on how the surface is oriented. No matter which way you orient the surface, the force that is acting on that surface will always be perpendicular. Consequently, pressure is essentially just a proportionality between force and area that has no bearing on direction.

To answer this question, it's useful to think of this conceptually as the balloon being "submerged" in the atmosphere. Remember, buoyant forces refer to the force caused by the displacement of a fluid, and both liquids and gasses count as fluids. Thus, in this case, the balloon can be thought of as submerged in a fluid in the same way that any object can be submerged in a liquid such as water.

We're trying to look for the point at which the balloon stops rising. This is analogous to a situation in which the balloon is "floating" in the fluid. In such a case, the upward and downward forces are equal and balance each other out.

So now, we need to identify the upward and downward forces acting on the balloon and set them equal to each other. The upward force will be the buoyant force due to the displacement of air from the atmosphere. The downward force will be the weight of the balloon.

We can further rewrite the mass of the balloon in terms of its weight and density.

Canceling out the gravity term, we obtain:

Next, it's important to realize that the volume of atmosphere being displaced is exactly the same as the volume of the balloon, since it is the balloon that is causing the displacement of air in the atmosphere. Thus:

Furthermore, because these two values are equal, we can cancel them out in the above expression to obtain the following.

Here, we have shown that once the density of the balloon becomes equal to that of the surrounding atmosphere, it will cease rising.

In this question, we're told that three containers of different shapes are filled with water. We're also told that each container has the same depth of water.

To find the pressure at the bottom of any of the containers, we'll need to remember the equation for pressure.

Since each of the containers is open to the atmosphere, we can disregard the term for atmospheric pressure in the above expression. Therefore, the absolute pressure becomes the gauge pressure.

Also, since each container is filled with water, the density of the fluid in each container is identical. Moreover, the depth we are considering for each container is also the same. Therefore, the pressure at the bottom of each container is exactly the same.

We can calculate the total pressure on the ball from the given force and radius:

Where for a sphere:

There are two pressures that combine to the total pressure on the ball: hydrostatic and atmospheric.

Since the ball is submerged in the ocean, we know that the surface of the water is at sea level and thus has a pressure:

Also, we can calculate the hydrostatic pressure with the following expression:

Where density is of the ocean and the height is the depth of the ball. Plugging this into our last expression, we get:

Rearranging for height:

To determine the hydrostatic pressure, we need to know the depth of the ship at time t = 12s. We can determine this with the following expression:

Then using the expression for hydrostatic pressure:

The following formula on pressure and area is used:

We substitute our known values and solve for F2 to obtain the output force:

Therefore the correct answer is of force.

The formula for pressure is given as:

Where is pressure in pascals, is force in newtons, and is area in meters squared. By substituting our known values we can solve for pressure:

Therefore the correct answer is

Given the hydrostatic pressure, we can calculate the depth that this occurs at:

Rearranging for height:

Plugging in our values, we get:

We can then use the following kinematics equation to determine how much time has passed:

If we designate the downward direction as positive and plugging in values to the kinematics equation, we get:

Rearranging, we get:

Since we can't have a negative time, the first one is the answer.

The total pressure on the ball includes both hydrostatic and atmospheric pressure:

We are given the atmospheric pressure, so we just need to determine the hydrostatic pressure using the following expression:

Plugging in values:

Therefore,

Now to determine the force on the ball, we need it's surface area. For a sphere:

Then,

We will use the expression for hydrostatic pressure for this problem:

In this scenario, the pressures from each material are additive. Therefore:

Plugging in expressions:

Plugging in values:

Note how we used 0.9m for the height of mercury since we are asked for the pressure at a height of 0.5m.

In this question, we're shown a hydraulic lift. A lift such as this functions to transmit a smaller force into a larger force via an incompressible liquid. From the diagram shown, we're asked to determine the relative values of the pressure and force on the right and left sides.

A hydraulic lift is able to transmit a small force into a larger force due to the liquid it contains being unable to compress. What this means is that when a force is applied to a given area of the liquid, this pressure is transmitted to all parts of the liquid. Therefore, at any given point, the pressure at all points within the lift will be equal. Thus, we can rule out the answer choices that list pressure.

But what about the forces? If the pressure is the same everywhere in an incompressible liquid, does this also mean the force is the same everywhere? The answer is no. Since pressure depends on the ratio of force to area, equal pressure does not mean equal force, because the area on which the force is acting must be taken into account. We can show this with the following expressions.

As can be seen from the above expression, the output force is greater than the input force by a factor equal to the ratio of the areas on which each force acts. In other words, to maintain constant pressure, a greater area will mean a greater force. Thus, from the diagram, we can see that the force on the right () will be greater than the force on the left ().

Pressure is force divided by area. The force Paul applies is . The rest is just algebra.