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statuscvo [17]
3 years ago
9

What are two processes that transfer water into the atmosphere

Physics
1 answer:
Ira Lisetskai [31]3 years ago
4 0
Water enters the atmosphere through evaporation, transpiration, excretion and sublimation: Transpiration is the loss of water from plant

Hope this helped you! :D
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Determine the lamp wattage required to obtain an illumination level of 50 f-c over a 100 ft2 area if a fixture is used with a CU
Bezzdna [24]

Lamp Wattage is utilized with a CU of 0.75 and 80 percent of the available light reaches the work surface, the remaining 20 percent is absorbed by walls and other objects in the space, resulting in an illumination level of 50 f-c over a 100 ft2 area

The term "lumen" refers to the emission of "luminous flux," which is a measurement of the total amount of visible light emitted by a source in a certain amount of time. The illumination level over a 100 ft2 area will be 50 f-c Lamp Wattage with a CU of 0.75 used, with 80 percent of the available light reaching the work surface and the remaining 20 percent being absorbed by walls and other objects in the room.
Total Lamp Wattage = Number of Lamps X Wattage of Each Lamp.
The fixtures' total wattage is 2 x 32, or 64 watts.
Lumen per Fixtures equals Lumen Efficiency (Lumen per Watt) times the Watt of each Fixture
85x 64 = 5440 lumens per fixture.

To learn more about lamp wattage please visit -
brainly.com/question/27154431
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3 0
2 years ago
A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the
fredd [130]

Answer:

  t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material  = 1.375

we know the equation

       c = ν λ

where ν is the frequency of the wave

           c is the speed of light

   \nu= \dfrac{c}{\nu\lambda}

   \nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}

       ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

        t = \dfrac{\lambda}{4\mu_{material}}

        t = \dfrac{522 \times 10^{-9}}{4\times 1.375}

              t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

7 0
3 years ago
PLEASE HELP : What happens in obese mice? (Physiology)
irina1246 [14]

Answer and

Explanation:

The gut microbiota has recently emerged as an important, and previously unappreciated, player in host physiology (1). In particular, the gut microbiota contributes to a variety of physiological and pathophysiological processes in the host including immune disorders (2–4), atherosclerosis (5), irritable bowel syndrome (6, 7), blood pressure regulation (8), and chronic kidney disease (9, 10). Bacteria residing in the human gut are an important component of human physiology: the total wet weight of gut microbes in the human has been estimated to be 175 g–1.5 kg (11, 12), and the cells of the microbiota outnumber human cells by 10:1 (1). These bacteria interact with the immune system of the host (13), and secrete a variety of metabolites, which enter host circulation and can affect a variety of physiological parameters (8, 14), reviewed in Ref. (15). In fact, metabolites produced by the gut microbiota have been found to play key roles in renal disease (16), blood pressure regulation (8), and immune disorders (2–4). Therefore, just as we consider the genetic background of an animal or an individual to be an important contributing factor to their physiology, so too must we consider the genetic background of the microbiota associated with that animal.

Gut microbiota vary greatly amongst laboratory animals, and these differences result in notable differences in experimental results. Mice of the same strain from different vendors have different microbiota profiles (17), and similarly, the same mice housed at different institutions have different microbiota profiles (18, 19). Conversely, inoculating two different inbred mouse strains with the same gut bacteria leads to differences in host gene expression between the two mouse strains (20). Clearly, there is a complex interplay between the genetics of the microbiota and that of the host organism, which has only recently begun to be appreciated.

Go to:

Gut Microbiota as an Experimental Parameter

Examples in the literature have highlighted the important and unexpected ways in which gut microbiota can affect a variety of experimental parameters. In a series of studies, Vijay-Kumar et al. (13, 21) reported that although TLR5 null animals initially had a colitis phenotype, when these mice were “rederived” and their gut microbiota altered, the colitis phenotype was greatly attenuated, and instead the null animals exhibited metabolic syndrome. In addition, Lathrop et al. put forward a model by which T-cells are educated not only by self/non-self mechanisms, but also by microbiota-derived “non-self” antigens (22). Accordingly, they found that the presence or absence of microbiota determined whether T cells would induce colitis in mice. Finally, Yang et al. reported that when the same knockout mice were housed at two different institutions, they had markedly different microbiota profiles – and the mice at one institution (MIT) were quite susceptible to colitis, whereas mice at the other institution (MHH) failed to develop any significant pathology under the same conditions (19). Unequivocally, altering gut microbiota – even by housing animals at different institutions – can have dramatic effects on the phenotype observed.

Go to:

Gut Microbiota and Obesity and Diabetes

It is important to note that not only can microbiota affect host physiology, but the gut microbiota are not necessarily stable over time. Rather, gut microbiota can change or shift as a result of experimental manipulation (in animals) or changes in lifestyle or nutrition (in humans). It is now appreciated that there are “shifts” in microbiota that occur in obesity in mice, rats, and humans (23–26). In one study, Turnbaugh et al. (25) examined human female twin pairs concordant for leanness or obesity, and found that obesity was associated with phylum-level changes in microbiota.

7 0
3 years ago
An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
jenyasd209 [6]

Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

7 0
3 years ago
What is scientific learning​
bixtya [17]
Hello here

It means that a person has the ability to describe, explain, and predict natural phenomena.
4 0
2 years ago
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