1).If we changed our speed limit signs to metric, what wouldprobably replace 55m
ID: 1666427 • Letter: 1
Question
1).If we changed our speed limit signs to metric, what wouldprobably replace 55mi/h? (Pleas round your answer to the nearest 1km/h? ------km/h 2). A jogger jogs around a circlar track with a diameter of225 m in 16 mintutes. What was the jogger's average speed in m/s^2?----- m/s 3). Joe Cool drives to see his girlfriend who attends anothercollege. (a) The 142 km trip takes 2.0h. what was the average speed forthe trip? ---km/h (b) In 4.0 more seconds, the car is going 92ft/s (63mi/h).what is the car's average acceleration for this timeperiod?---ft/s^2 (c) The car then slows to 69 ft/s (47 mi/h) in 3.0s. What isthe average acceleration for this time period?------ft/s^2 (d) What is the overall average acceleration for the totaltime? -----ft/s^2 4). A rectangle container measuring 17cm*52cm*66cm is filledwith water. what is the mass of this volume of water in kilogramsand grams? 1).If we changed our speed limit signs to metric, what wouldprobably replace 55mi/h? (Pleas round your answer to the nearest 1km/h? ------km/h 2). A jogger jogs around a circlar track with a diameter of225 m in 16 mintutes. What was the jogger's average speed in m/s^2?----- m/s 3). Joe Cool drives to see his girlfriend who attends anothercollege. (a) The 142 km trip takes 2.0h. what was the average speed forthe trip? ---km/h (b) In 4.0 more seconds, the car is going 92ft/s (63mi/h).what is the car's average acceleration for this timeperiod?---ft/s^2 (c) The car then slows to 69 ft/s (47 mi/h) in 3.0s. What isthe average acceleration for this time period?------ft/s^2 (d) What is the overall average acceleration for the totaltime? -----ft/s^2 4). A rectangle container measuring 17cm*52cm*66cm is filledwith water. what is the mass of this volume of water in kilogramsand grams?Explanation / Answer
1 km = .621 mi or 1 mi = 1.61 km (remember 1 km˜ .62 mi) 1) 55 mi / hr * 1.61 km / mi = 88.5 km / hr = 88 km / hr (if .5 round to even number) 2) C = d = 707 m circumference oftrack a) v = 707 m / 16 min = 44.2 m / min =44.2 m / min / (60 sec / min) = .74 m/sec b) vf = 3) vavg = s / t = 142 km / 2 hr = 71 km / hr= 71 km / hr * .62 mi / km = 44 mph An easy conversion factor is 60 mph = 88 ft / sec 44 mph = 44 / 60 * 88 = 64.6 ft / sec 63 mph = 63 / 60 * 88 = 92.4 ft / sec (vf - vavg) / t = (92.4 -64.6) / 4 = 7 ft / sec2 47 mph = 47 / 60 * 88 = 69 ft / sec (vf - vavg) / t = (69 - 92) / 3 =- 7.7 ft / sec2 d) aavg = (vf - vi) / t = (69- 65) / 7 = -.57 ft/sec2 (final speed - initial speed) / time 4) V = 17 * 52 * 66 = 5.83 * 10E4 cm3 M = 5.83 * 10E4 cm3 * 1 gm / cm3 = 5.83 * 10E4 gm =58.3 kg M = 5.83 * 10E4 cm3 * 1 gm / cm3 = 5.83 * 10E4 gm =58.3 kgRelated Questions
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