Anything less that 7, would be a base.
Frequency = speed of light
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wavelength
= 3 x 10^8
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344 x 10^-9
= 8.72 x 10^14 Hz.
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Two [hydrogen] atoms and one [oxygen] atom
Answer:
The best existing methods of control—artemisinin-based drug treatment and insect control with chemical sprays and treated bednets—can reduce the burden of disease substantially, and can even eliminate the disease in some regions,
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Answer:In general, yes. Usually, you talk about the reflectivity/emissivity of a surface rather than an object. Typically, you’d find a coefficient of emissivity (since emission and absorption are normally equivalent processes in reverse) for the surface, ideally as a function of wavelength and incident angle. Then you apply that coefficient to all light striking the object.
Assuming an opaque object, reflectivity (call it R) has a simple relationship to emissivity (call it E): R=1-E. You can measure reflectivity with a calibrated light source and light sensor, for example. It’s a little trickier, but you can also measure emissivity through techniques like calorimetry, where you measure how much something heats up to figure out how much energy was deposited; again, a calibrated light source can be used to direct a certain amount of power onto a test object, and the heating tells you how much power is retained. You can also get emissivity by heating an object and observing how much power it emits by blackbody radiation.
To be really thorough, you might also want to measure transmissivity, in case the object isn’t opaque. If we call transmissivity “T,” we really have to write R+E+T=1 (which just says that all the incident light has to either reflect, absorb, or pass through).
If the object is something celestial, of course, it’s harder to use these methods, but not impossible. For example, we can measure how much light the full moon reflects, and knowing how much light hits it from the sun, we can find the reflectivity; in principle, we could use our knowledge of how much the surface of the moon is heated by the sun to find the emissivity, as well. For that kind of calculation, it’s important to know how large and how distant an object is, to figure out how much of its reflected light makes it back to you.
Explanation:
