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You are designing a diving bell to withstand the pressure of seawater at a depth of 250 \(\mathrm{m}\) . (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth. \((b)\) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 \(\mathrm{cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You can ignore the small variation of pressure over the surface of the window.)

Short Answer

Expert verified
(a) Gauge pressure is 2.45 MPa. (b) Net force on the window is approximately 166,048 N.

Step by step solution

01

Calculate the Water Pressure at Depth

To find the gauge pressure at 250 m depth, we use the formula for pressure in a fluid, \( P = \rho g h \), where \( \rho \) is the density of water (~1000 kg/m³), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the depth (250 m). Plugging in the values, we compute:\[P = (1000 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(250 \text{ m}) = 2,450,000 \text{ Pa}\]Thus, the gauge pressure at 250 m depth is 2.45 MPa.
02

Find Diameter in Meters and Calculate Area of Window

The diameter of the circular window is given as 30.0 cm. First, convert this to meters by dividing by 100:\[ \text{Diameter in meters} = \frac{30.0 \text{ cm}}{100} = 0.30 \text{ m} \]Next, calculate the area of the circular window using the area formula \( A = \pi r^2 \), where \( r = \frac{0.30}{2} = 0.15 \) meters.\[A = \pi (0.15)^2 \approx 0.0707 \text{ m}^2\]Therefore, the area of the window is approximately 0.0707 square meters.
03

Calculate Net Force on the Window

To find the net force on the glass window, calculate the force due to water pressure and subtract the force due to the air inside. Since the pressure inside the diving bell equals the atmospheric pressure (~101,325 Pa), the net gauge pressure is: \[ P_{\text{net}} = P_{\text{water}} - P_{\text{air}} = 2,450,000 \text{ Pa} - 101,325 \text{ Pa} = 2,348,675 \text{ Pa} \]Multiply the net pressure by the area of the window to find the net force:\[F_{\text{net}} = P_{\text{net}} \times A = 2,348,675 \text{ Pa} \times 0.0707 \text{ m}^2 \approx 166,048 \text{ N}\]Therefore, the net force on the window is approximately 166,048 N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauge Pressure
In fluid mechanics, gauge pressure refers to the pressure of a fluid measured relative to the ambient atmospheric pressure. It does not include atmospheric pressure itself. This type of pressure is the most straightforward measure in applications like scuba diving or underwater exploration.
To calculate gauge pressure at a specific depth in a fluid, such as water, you use the formula:
  • \( P = \rho g h \)
  • Where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth.
In our exercise, with water at a density of approximately 1000 kg/m³ and at a depth of 250 meters, the gauge pressure is calculated to be 2.45 MPa, indicating significant pressure experienced underwater, aside from atmospheric pressure.
Net Force
The net force is essentially the resultant force acting upon an object when all the individual forces are combined. Underwater, the net force on an object like a window of a diving bell is determined by considering the difference between the pressure exerted by the water from the outside and the pressure inside the bell.

To calculate it:
  • First, determine the net pressure: \( P_{\text{net}} = P_{\text{water}} - P_{\text{air}} \)
  • Then, multiply this by the window area to obtain the force: \( F_{\text{net}} = P_{\text{net}} \times A \)
Using given values, the net force in our example is approximately 166,048 N, acting inward on the window. This demonstrates the need for robust materials and design in underwater equipment to withstand such forces.
Circular Window Area
The circular window area is crucial to understanding how much pressure acts on each point of the window. The area needs to be calculated first before determining the force. For a circular window, the formula to find the area is:
  • \( A = \pi r^2 \)
  • Where \( r \) is the radius of the circle.
In the exercise, the window's diameter was given as 30 cm, which was converted into meters, resulting in a radius of 0.15 meters. Plugging these into the area formula results in approximately 0.0707 square meters. Calculating this accurately is essential for subsequent force calculations.
Underwater Pressure
Underwater pressure is a crucial concept in fluid mechanics, as it affects everything from diver safety to the integrity of submersible design. The pressure underwater increases significantly with depth due to the weight of the water above.
This pressure can be quantified using the previously mentioned equation \( P = \rho g h \).

The pressure differential created by the water column is the primary factor designers must consider when constructing underwater vehicles or habitats like diving bells. In the exercise, the calculation shows a substantial pressure of 2.45 MPa at 250 m depth, indicating that both the structure and material strength are vital for safe underwater operations.

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Most popular questions from this chapter

A piece of wood is 0.600 \(\mathrm{m}\) long, 0.250 \(\mathrm{m}\) wide, and 0.080 \(\mathrm{m}\) thick. Its density is 600 \(\mathrm{kg} / \mathrm{m}^{3}\) . What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

A rock with mass \(m=3.00 \mathrm{kg}\) is suspended from the roof of an elevator by a light cord. The rock is totally immersed in a bucket of water that sits on the floor of the elevator, but the rock doesn't touch the bottom or sides of the bucket. (a) When the elevator is at rest, the tension in the cord is 21.0 \(\mathrm{N}\) . Calculate the volume of the rock. ( b) Derive an expression for the tension in the cord when the elevator is accelerating upward with an acceleration of magnitude a. Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) upward. (c) Derive an expression for the tension in the cord when the clevator is accelerating downwand with an acceleration of magnitude \(a\) . Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) downward. (d) What is the tension when the elevator is in free fall with a downward acceleration equal to \(g ?\)

What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 \(\mathrm{m} ?\) (Assume that the mains have a much larger diameter than the fire hose.)

Untethered helium balloons, floating in a car that has all the windows rolled up and outside air vents closed, move in the direction of the car's acceleration, but loose balloons filled with air move in the opposite direction. To show why, consider only the horizontal forces acting on the balloons. Let \(a\) be the magnitude of the car's forward acceleration. Consider a horizontal tube of air with a cross-sectional area \(A\) that extends from the windshield, where \(x=0\) and \(p=p_{0},\) back along the \(x\) -axis. Now consider a volume element of thickness \(d x\) in this tube. The pressure on its front surface is \(p\) and the pressure on its rear surface is \(p+d p\) . Assume the air has a constant density \(\rho\) . (a) Apply Newton's second law to the volume element to show that \(d p=\rho a d x\) . (b) Integrate the result of part (a) to find the pressure at the front surface in terms of \(a\) and \(x\) . (c) To show that considering \(\rho\) constant is reasonable, calculate the pressure difference in atm for a distance as long as 2.5 \(\mathrm{m}\) and a large acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (d) Show that the net horizontal force on a balloon of volume \(\dot{V}\) is \(\rho V a\) . (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text { bal }} )\) is \(\left(\rho / \rho_{\text { bal }}\right) a,\) so that the acceleration relative to the car is \(a_{n e 1}=\left[\left(\rho / \rho_{\text { bel }}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rel}}\) in part (e) to explain the movement of the balloons.

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