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What radius must a water drop have for the difference between the inside and outside pressures to be 0.0200 atm? The surface tension of water is 72.8 dynes/cm.
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Physics 101 Mechanics
Chapter 13
Fluid Mechanics
Temperature and Heat
Rutgers, The State University of New Jersey
University of Michigan - Ann Arbor
University of Sheffield
McMaster University
Lectures
03:45
In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.
09:49
A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.
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okay. And this problem were asked what radius of water droplet must have in order that the excess pressure is equal to 0.2 atmospheres. So first off, we want Teo Express this pressure difference of 0.2 atmospheres in terms of past calls and come the conversion factors that one atmosphere is equal to one point Owen three times 10 to the five past calls. So this expression here will give us the the excess pressure in the units that we want and the formula that relates the excess pressure to the surface Tension and the radius is this equation here. Delta P equals to gamma over our and were given that the surface tension of water gamma is equal to 72.8 dynes per centimeter and one dime per centimeter is equal to 10 to the negative three moons per meter. So all together at the surface tension express them units that we want is given by this If so, Lastly, if we want to find the radius are what we want to do is solve this equation for our we have our equals to gamma over Delta P, which is equal to two Times 78 Excuse me, 72.8 times 10 to the negative three. Divided by Delta P, which is 0.0 too. Times 1.13 times 10 to the five.
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