COLLEGE PHY2053 W/MODIFIED ACCESS>BI<
16th Edition
ISBN: 9781323515303
Author: Knight
Publisher: PEARSON C
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Chapter 28, Problem 34MCQ
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T1. Calculate what is the received frequency when the car drives away from the radar antenna at a speed v of a) 1 m/s ( = 3.6 km/h), b) 10 m/s ( = 36 km/h), c) 30 m /s ( = 108 km/h) . The radar transmission frequency f is 24.125 GHz = 24.125*10^9 Hz, about 24 GHz. Speed of light 2.998 *10^8 m/s.
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Chapter 28 Solutions
COLLEGE PHY2053 W/MODIFIED ACCESS>BI<
Ch. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQ
Ch. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 16CQCh. 28 - Prob. 17CQCh. 28 - Prob. 18CQCh. 28 - Prob. 19CQCh. 28 - Prob. 20CQCh. 28 - Prob. 21CQCh. 28 - Prob. 22CQCh. 28 - Prob. 23CQCh. 28 - Prob. 24CQCh. 28 - Prob. 25CQCh. 28 - Prob. 26CQCh. 28 - Prob. 27CQCh. 28 - Prob. 28MCQCh. 28 - Prob. 29MCQCh. 28 - Prob. 30MCQCh. 28 - Prob. 31MCQCh. 28 - Prob. 32MCQCh. 28 - Prob. 33MCQCh. 28 - Prob. 34MCQCh. 28 - Prob. 35MCQCh. 28 - Prob. 36MCQCh. 28 - Prob. 37MCQCh. 28 - Prob. 38MCQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49PCh. 28 - Prob. 50PCh. 28 - Prob. 51PCh. 28 - Prob. 52PCh. 28 - Prob. 53PCh. 28 - Prob. 54PCh. 28 - Prob. 55PCh. 28 - Prob. 56PCh. 28 - Prob. 57PCh. 28 - Prob. 58GPCh. 28 - Prob. 59GPCh. 28 - Prob. 60GPCh. 28 - Prob. 61GPCh. 28 - Prob. 62GPCh. 28 - Prob. 63GPCh. 28 - Prob. 64GPCh. 28 - Prob. 65GPCh. 28 - Prob. 66GPCh. 28 - Prob. 67GPCh. 28 - Prob. 68GPCh. 28 - Prob. 69GPCh. 28 - Prob. 70GPCh. 28 - Prob. 71GPCh. 28 - Prob. 72GPCh. 28 - Prob. 73GPCh. 28 - Prob. 74GPCh. 28 - Prob. 75GPCh. 28 - Prob. 76GPCh. 28 - Prob. 77GPCh. 28 - Prob. 78GPCh. 28 - Prob. 79MSPPCh. 28 - Prob. 80MSPPCh. 28 - Prob. 81MSPPCh. 28 - Prob. 82MSPPCh. 28 - Prob. 83MSPPCh. 28 - Prob. 84MSPPCh. 28 - Prob. 85MSPPCh. 28 - Prob. 86MSPPCh. 28 - Prob. 87MSPPCh. 28 - Prob. 88MSPPCh. 28 - Prob. 89MSPPCh. 28 - Prob. 90MSPPCh. 28 - Prob. 91MSPPCh. 28 - Prob. 92MSPP
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- No Chatgpt pleasearrow_forward3. A measurement taken from the UW Jacobson Observatory (Latitude: 47.660503°, Longitude: -122.309424°, Altitude: 220.00 feet) when its local sidereal time is 120.00° makes the following observations of a space object (Based on Curtis Problems 5.12 + 5.13): Azimuth: 225.00° Azimuth rate: 2.0000°/s. Elevation: 75.000° Elevation rate: -0.5000°/s Range: 1500.0 km Range rate: -1.0000 km/s a. What are the r & v vectors (the state vector) in geocentric coordinates? (Answer r = [-2503.47 v = [17.298 4885.2 5.920 5577.6] -2.663]) b. Calculate the orbital elements of the satellite. (For your thoughts: what type of object would this be?) (Partial Answer e = 5.5876, 0=-13.74°) Tip: use Curtis algorithms 5.4 and 4.2.arrow_forwardConsider an isotope with an atomic number of (2(5+4)) and a mass number of (4(5+4)+2). Using the atomic masses given in the attached table, calculate the binding energy per nucleon for this isotope. Give your answer in MeV/nucleon and with 4 significant figures.arrow_forward
- A: VR= 2.4 cm (0.1 V/cm) = 0.24 V What do Vector B an C represent and what are their magnitudesarrow_forward4. Consider a cubesat that got deployed below the ISS and achieved a circular orbit of 410 km altitude with an inclination of 51.600°. What is the spacing, in kilometers, between successive ground tracks at the equator: a. Ignoring J2 (Earth's oblateness) effects b. Accounting for J2 effects c. Compare the two results and comment [Partial Answer: 35.7km difference]arrow_forwardplease solve and explainarrow_forward
- Two ice skaters, both of mass 68 kgkg, approach on parallel paths 1.6 mm apart. Both are moving at 3.0 m/sm/s with their arms outstretched. They join hands as they pass, still maintaining their 1.6 mm separation, and begin rotating about one another. Treat the skaters as particles with regard to their rotational inertia. a) What is their common angular speed after joining hands? Express your answer in radians per second. b) Calculate the change in kinetic energy for the process described in a). Express your answer with the appropriate units. c) If they now pull on each other’s hands, reducing their radius to half its original value, what is their common angular speed after reducing their radius? Express your answer in radians per second. d) Calculate the change in kinetic energy for the process described in part c). Express your answer with the appropriate units.arrow_forwardPlease solve and explainarrow_forwardNo Chatgpt pleasearrow_forward
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