Calculate the reverberation time in seconds at 500 Hz for a conference room that is 13.3 feet wide, 22.2 feet long, and 11.1 feet high. It has a carpeted floor, curtains on both long sides, plasterboard short sides and a plasterboard ceiling. The conference room is also occupied by 20 people in comfortable upholstered seats.
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Calculate the reverberation time in seconds at 500 Hz for a conference room that is 13.3 feet wide, 22.2 feet long, and 11.1 feet high. It has a carpeted floor, curtains on both long sides, plasterboard short sides and a plasterboard ceiling. The conference room is also occupied by 20 people in comfortable upholstered seats.
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- Please asap2.5 Represent and reason A 2.0-kg cart attached to a horizontal spring vibrates on a low-friction track (similar to the situation shown in Activity 1.3). The cart's displacement-versus-time is described by: 2 x = (0.20m) cos ( |COS 2.0s What is the meaning of each number in the function above. Determine the period and the amplitude of the motion. The positive direction of the x-axis is to the right. Work with your group to determine the cart's position at t = 0, t = 0.5 s, t= 1.0 s, and t= 1.5 s. Discuss: What is special about the chosen times? (Hint: What fraction of a period is 0.5 s, 1.0 s, and 1.5 s?)Both parts are very confusing for me could I get a broken down response for all steps for both a and b
- The mass is pulled to the right a distance of 0.2 m and released. Rank the following spring–mass combinations according to their oscillation periods from longest to shortest. If any combinations have the same period, give them the same rank. You should assume that there is no friction between the mass and the horizontal surface. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) k = 0.4 N/m; m = 0.50 kg k = 0.4 N/m; m = 1.00 kg k = 0.4 N/m; m = 2.00 kg k = 0.8 N/m; m = 0.50 kg k = 0.8 N/m; m = 1.00 kgWhy is Logger Pro set up to report the time between every other blocking of the Photogate? Why not the time between every block? Using either Graphical Analysis or graph paper, plot a graph of pendulum period vs. amplitude in degrees. Scale each axis from the origin (0,0). Does the period depend on amplitude? Explain. Using either Graphical Analysis or graph paper, plot a graph of pendulum period T vs. lengthl. Scale each axis from the origin (0,0). Does the period appear to depend on length? Using either Graphical Analysis or graph paper, plot the pendulum period vs. mass. Scale each axis from the origin (0,0). Does the period appear to depend on mass? Do you have enough data to answer conclusively? To examine more carefully how the period T depends on the pendulum length l, create the following two additional graphs of the same data: T 2 vs. l and T vs. l2 . Of the three period-length graphs, which is closest to a direct proportion; that is, which plot is most nearly a straight line…Select the best answer for the question. 10. Two recording devices are set 3,800 feet apart, with the device at point A to the west of the device at point B. At a point on a line between the devices, 400 feet from point B, a small amount of explosive is detonated. The recording devices record the time the sound reaches each one. How far directly north of site B should a second explosion be done so that the measured time difference recorded by the devices is the same as that for the first detonation? O A. 1,857.42 feet O B. 5,415.26 feet O C. 906.67 feet O D. 1,882.03 feet O Mark for review (Will be highlighted on the review page)
- The mass is pulled to the right a distance of 0.2 m and released. Rank the following spring–mass combinations according to their oscillation periods from longest to shortest. If any combinations have the same period, give them the same rank. You should assume that there is no friction between the mass and the horizontal surface. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) k = 0.3 N/m; m = 0.50 kg k = 0.3 N/m; m = 1.00 kg k = 0.3 N/m; m = 2.00 kg k = 0.6 N/m; m = 0.50 kg k = 0.6 N/m; m = 1.00 kgTwo strings both vibrate at exactly 764 Hz. The tension in one of them is then increased slightly. As a result, 6 beats per second are heard when both strings vibrate. What is the new frequency of the string that was tightened? Calculate to one decimal.For Parts I and II, use the pendulum equation and solve to predict the length needed for a pendulum with a period of 1 s and 2 s respectively. Use a piece of string (or thread, dental floss, shoe lace, etc.) and a steel nut (or washer or something small but with enough mass to weigh down the string) to build each pendulum. Time each pendulum for 30 periods and then find the average time for one period. Remember, a period is the time to complete one cycle of motion, out and back. Tape each pendulum up in perhaps a doorway where it is stationary and has room to swing. Be precise with your measuring and timing. Show your work. Part I Given: T = 1.00 s l =? Part II Given: T = 2.00 s l =?
- Calculate the amplitude A of a progressive wave with frequency f= 99 Hz and wavelength > =10 m if its displacement at the point x = 0.1 m at the time t= 10 s is 0.3 m. Provide your answer in SI units. Enter your answer in the box below. Answer: Choose...You will fire the spring gun 3 times from the first detent and measure the change in height of the (pendulum + ball) for each shot. Write the equation for the change in height of the first shot.Two cars have the same single-frequency horn. When one is at rest and the other is moving toward it at 15 m/s, the dirver at rest hears a beat frequency of 4.5 Hz. What is the frequency of the horns? Please sketch the situation, define all variables. Thank you