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Chapter 11: Problem 22

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance \(h,\) as the drawing shows. Using \(1.013 \times 10^{5}\) Pa for the atmospheric pressure and \(1200 \mathrm{kg} / \mathrm{m}^{3}\) for the density of the sauce, find the absolute pressure in the bulb when the distance \(h\) is \((a) 0.15 \mathrm{m}\) and \((b) 0.10 \mathrm{m}.\)

Short Answer

Expert verified
99534.2 Pa for \( h = 0.15 \text{ m} \); 100122.8 Pa for \( h = 0.10 \text{ m} \).

Step by step solution

01

Understanding the Scenario

We have a meat baster which works by creating a pressure difference that sucks sauce up the tube when the bulb is squeezed. The height to which the sauce rises, denoted by \( h \), depends on this pressure difference.
02

Identify the Known Values

We are given the atmospheric pressure as \( P_0 = 1.013 \times 10^5 \text{ Pa} \) and the density of the sauce \( \rho = 1200 \text{ kg/m}^3 \). We need to find the absolute pressure in the bulb for \( h = 0.15 \text{ m} \) and \( h = 0.10 \text{ m} \).
03

Use the Hydrostatic Pressure Equation

The pressure difference caused by the sauce column can be determined with the hydrostatic pressure equation: \[ P = P_0 - \rho g h \]where \( P \) is the pressure in the bulb, \( g \) is the gravitational acceleration \( 9.81 \text{ m/s}^2 \), and \( h \) is the height of the sauce column.
04

Calculate Pressure in the Bulb for h = 0.15 m

Substitute the known values for \( h = 0.15 \text{ m} \): \[ P = 1.013 \times 10^5 \text{ Pa} - 1200 \times 9.81 \times 0.15 \]Calculate:\[ P = 1.013 \times 10^5 \text{ Pa} - 1765.8 \text{ Pa} \] \[ P = 99534.2 \text{ Pa} \]
05

Calculate Pressure in the Bulb for h = 0.10 m

Now, substitute the known values for \( h = 0.10 \text{ m} \): \[ P = 1.013 \times 10^5 \text{ Pa} - 1200 \times 9.81 \times 0.10 \]Calculate: \[ P = 1.013 \times 10^5 \text{ Pa} - 1177.2 \text{ Pa} \] \[ P = 100122.8 \text{ Pa} \]
06

Solution Summary

For \( h = 0.15 \text{ m} \), the pressure in the bulb is \( 99534.2 \text{ Pa} \), and for \( h = 0.10 \text{ m} \), the pressure in the bulb is \( 100122.8 \text{ Pa} \).

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is a critical concept in understanding the meat baster scenario. It is the pressure exerted by the weight of the atmosphere on objects at the Earth's surface. When using a meat baster, atmospheric pressure plays a role in how fluids are sucked up.
In our example, the atmospheric pressure is given as \(1.013 \times 10^{5}\) Pascals (Pa). This value acts as the baseline pressure, which opposes the pressure changes inside the bulb.
  • When the bulb is squeezed, the pressure inside decreases below atmospheric pressure. This difference creates a mechanism for the basting sauce to rise up the tube.
  • The greater the atmospheric pressure, the greater the force available to push the sauce up when the internal pressure decreases.
Understanding atmospheric pressure helps predict how high the sauce can rise based on the changes in internal pressure relative to it.
Density of Liquids
Density is another vital concept when dealing with hydrostatic pressure and fluid mechanics. The density of a liquid is its mass per unit volume, typically measured in kilograms per cubic meter (kg/m鲁). In the meat baster problem, the sauce density is given as 1200 kg/m鲁.
This value is crucial in understanding how high the sauce will rise.
  • The denser the liquid, the more mass is in a given volume, which affects the pressure it exerts.
  • Higher density means that for the same volume, the liquid will exert greater force due to gravity, resulting in more pressure at the base of a column of that liquid.
When calculating the pressure difference, knowing the density allows you to determine how much pressure is added by the sauce column's weight. Ultimately, this affects how the internal pressure in the bulb competes with atmospheric pressure to lift the sauce.
Pressure Difference
Pressure difference is a core principle in the operation of fluid-related tools like a meat baster. It's the difference between two pressure points, such as the atmospheric pressure outside the bulb and the reduced pressure inside the bulb when squeezed.
Creating this pressure difference is what allows the basting sauce to rise. The basic relationship of pressure difference with height \(h\), density \(\rho\), and gravitational acceleration \(g\) is expressed by the hydrostatic equation: \[ \Delta P = \rho g h \] This pressure difference is subtracted from atmospheric pressure to find the pressure inside the bulb: \[ P = P_0 - \Delta P \]
  • Greater height \( h \) results in a larger pressure difference because more liquid trying to be pushed upwards against gravity translates to greater weight.
  • The pressure difference must overcome the atmospheric pressure to make the liquid rise, and this is possible because the pressure inside the bulb is lowered.
  • Both the distance \( h \) the sauce rises and the density \( \rho \) are fundamental in calculating \( \Delta P \), and thus ultimately determine the internal pressure of the bulb during use.

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Most popular questions from this chapter

One way to administer an inoculation is with a "gun" that shoots the vaccine through a narrow opening. No needle is necessary, for the vaccine emerges with sufficient speed to pass directly into the tissue beneath the skin. The speed is high, because the vaccine \(\left(\rho=1100 \mathrm{kg} / \mathrm{m}^{3}\right)\) is held in a reservoir where a high pressure pushes it out. The pressure on the surface of the vaccine in one gun is \(4.1 \times 10^{6}\) Pa above the atmospheric pressure outside the narrow opening. The dosage is small enough that the vaccine's surface in the reservoir is nearly stationary during an inoculation. The vertical height between the vaccine's surface in the reservoir and the opening can be ignored. Find the speed at which the vaccine emerges.

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A small crack occurs at the base of a 15.0 -m-high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

A person who weighs \(625 \mathrm{N}\) is riding a 98 -N mountain bike. Suppose that the entire weight of the rider and bike is supported equally by the two tires. If the pressure in each tire is \(7.60 \times 10^{5} \mathrm{Pa}\), what is the area of contact between each tire and the ground?

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Because of the aneurysm, the normal cross-sectional area \(A_{1}\) of the aorta increases to a value of \(A_{2}=1.7 A_{1} .\) The speed of the blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}\right)\) through a normal portion of the aorta is \(v_{1}=0.40 \mathrm{m} / \mathrm{s} .\) Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure \(P_{2}\) in the enlarged region exceeds the pressure \(P_{1}\) in the normal region.
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