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stiv31 [10]
3 years ago
12

What are air masses, and how do they work? NEED HELP ASAP!

Physics
2 answers:
Georgia [21]3 years ago
5 0
Also known as air-flow mattresses, alternating pressure mattresses relieve and redistribute pressure through a dynamic lying surface. Alternating pressure sore mattresses are assembled with a row of lateral air cells that constantly alternate. Hope it helps
nikitadnepr [17]3 years ago
3 0

Answer:

Need help? I'll give it to you!

Explanation:

Air masses are formed when air stagnates for long periods of time over a uniform surface. The characteristic temperature and moisture of air masses are determined by the surface over which they form. An air mass acquires these attributes through heat and moisture exchanges with the surface.

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What is the relationship between net force and balanced forces
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Which of the following situations describes a change that will result in a new kind of matter with different characteristics?
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2 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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