good evening those who are watching. Here is my question Ireally need some insit
ID: 1667201 • Letter: G
Question
good evening those who are watching. Here is my question Ireally need some insite on this question please. Modern oiltankers weigh over a half-million tons and have lengths of up to aquarter of a mile. Such massive ships require a distance of 5.0 km( about 3.0 mi) and a time of 19 min to come to a stop from a topspeed of 34 km/h. a) What is the magnitude of such a ship's average accelerationin m/s2 in coming to a stop? ___________m/s2 b) What is the magnitude of the ship's average velocity inm/s? ____________m/s c) whats the potential of a tanker runner aground? good evening those who are watching. Here is my question Ireally need some insite on this question please. Modern oiltankers weigh over a half-million tons and have lengths of up to aquarter of a mile. Such massive ships require a distance of 5.0 km( about 3.0 mi) and a time of 19 min to come to a stop from a topspeed of 34 km/h. a) What is the magnitude of such a ship's average accelerationin m/s2 in coming to a stop? ___________m/s2 b) What is the magnitude of the ship's average velocity inm/s? ____________m/s c) whats the potential of a tanker runner aground?Explanation / Answer
For the first two part, this should be sovle using the equation ofdisplacement and velocity. y=y0+v0*t+at^2/2 and v=v0+a*t In this case, y0 and v0 are both zero. so y=at^2/2 and v=at. a). average acceleration a=v/t so a=34km/h / 19min. b) similar to above. average velocity v=y/t v=5km/19min. c)Since the average acceleration of the tanker is .... (part a) We can assume that there is a force which is equal to F=ma act onthe ship. so the potential should be W=-F*s (using the conservation ofenergy, we have W+K=const , play with the equations abit and youwill come up with F*s=mv^2/2. so it is equal to -F*s+mv^2/2=0=const, mv^2/2=K so -Fs=W)
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