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Rutherford’s Model and мультфильмы для детей 6 летthe Electromagnetic SpectrumQuestion 1: What are the frequency and wavelength of an EM wave of energy 6.626 x 10-19 J?Answer: Frequency(f) = E/h = 1015 Hz. Wavelength(λ) = c/f = 3 x 108 / 1015 = 3 x 107 metresQuestion 2: What are the uses of X-rays?Answer: X-rays can be секс скачать used to detect medical ailments and bone deformities in medical diagnosis. They are useful for ionisation purposes also. 
Question 3: Can we use X-rays and gamma rays for broadcasting radio/TV/mobile signals?Answer: No, X-rays and gamma rays are short-range. Moreover, they are harmful and have penetrating дэдпул и росомаха смотреть бесплатно power on the matter with which they interact and can damage the tissues of living bodies.Question 4: Which ледниковый период фильм among the following is not a property of electromagnetic waves?1) Momentum2) Energy3) Pressure4) Heat energyAnswer:1) EM waves can impart momentum (and angular китайский секс momentum) to the material with which it interacts.
2) Electromagnetic waves carry energy. EM waves are the only waves able to carry energy across a vacuum.3) EM waves also exert pressure, which is shown by a radiometer. One side of the panels is black, the other side is white, and the panels spin due to the pressure differences under the light.4) EM waves do not carry heat energy, but any EM radiation can heat an object when it is absorbed.Question 5: A ray from the sun, passing through your kitchen window, hits a prism that casts a rainbow on the windowsill. Supposedly, there is a hand-held radiometer on the table. Now, you place the instrument on a specific colour of the rainbow with your eyes closed. When you open your eyes, you see that the radiometer measured the energy from that colour at 4.0 x 10-19 joules. If we take Planck’s constant of 6.6256 x 10-34 joules/sec, what possible colour did the instrument measure? How can we determine this?Answer: We should use the equation involving energy change, Planck’s constant, and frequency. Next, we need to figure out what we are solving for. In this problem, they ask for the possible colour that you measured. If we relate the energy with Planck’s constant, we can solve for frequency. We are given the energy, 4.0 x 10-19 joules, as well as Planck’s constant, 6.6256 x 10-34 joules/sec. Also, we are given the frequencies emitted by the visible spectrum, from red to violet. This problem is easy to solve now. If we solve for the frequency, we can then relate it to the energy emitted, measured in either sec-1.Let’s solve it.– What is the best equation to use? E = h* ƒ– Solve for the intended variable. E/ h = ƒ– Energy measured = 4.0 x 10-19 joulesPlanck’s constant = 6.6256 x 10-34 joulesваяна 2 смотреть онлайн/secVisible spectrum frequency(4.0 x 10-19 joules) / 6.6256 x 10-34 joules/sec = ƒ– Joules cancel out with joules, and one is left with sec-1, a frequency.Answer = 6.03 x 1014 sec-1This falls within the given visible spectrum frequencies. Being a high frequency on the visible spectrum, close to the frequencies emitted by blue colour, one could assume that you measured a colour close to blue, though cyan and green cannot be out of question as the given frequencies are a bit vague. It is closest to the green colour. Hence, we measured the green colour. |
