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4 and 5 exercises, calculations Student 1. In an LC circuit in which C 4.0 HF th

ID: 1571317 • Letter: 4

Question


4 and 5 exercises, calculations

Student 1. In an LC circuit in which C 4.0 HF the maximum potent difference across the capacitor during the oscillations is 17 v and the maximum cur through the inductor is 40 mA. (a) What is the inductance L? (b) what is the frequency of the (c How much time does it take for the charge on the capacitor to rise from zero to the half of its maximum value? 2. One type of hologram consists of bright and dark fringes produced on photographic film by interfering laser beams. If the hologram is illuminated with white light, the image will be reproduced multiple times in different pure colors at different sizes. A) explain why b) which color correspond to the largest and smallest images, and why? diffraction grating has 300 rulings mm, and a strong diffracted beam is noted at o 250. (a) What are the possible wavelengths of the incident light? (b) What colors are they? 4. An oven with an inside temperature To 317c is in a room having a temperature T 270c. There is a small opening of area 8.0 cm in one side of the oven. How much net power is transferred from the oven to the room? 5. The work needed to remove an electron from the surface of sodium is 2.2 ev. Find the How maximum wavelength of light that will cause photoelectrons to be emitted from sodium. big maximal velocity will have electrons if the surface of sodium will be illuminated with ultraviolet light (wavelength 425 nm)? 6. What is more dangerous, a radioactive material with a short halflife time or a long one?

Explanation / Answer

4) Net power transferred = k * A * T4

                                           = 5.67 * 10-8 * 8 * 10-4 * (590.154 - 300.154)

                                           = 5.1338 W

5) maximum wavelength of light =   (6.63 * 10-34 * 3 * 108)/(2.2 * 1.6 * 10-19)

                                                    = 565 nm

    Here, energy of electrons = (6.63 * 10-34 * 3 * 108)/(425 * 10-9) -   2.2 * 1.6 * 10-19

                                                                  =   1.16 * 10-19 J

=> velocity of electrons = sqrt(2 * 1.16 * 10-19/9.11 * 10-31)

                                         = 5.0464 * 105 m/s

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