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At a distance of 4,000 ft from the launch site, a spectator is observing a rocket being launched. If the rocket lifts off vertically and is rising at a speed of 600 ft/sec when it is at an altitude of 3,000 ft, how fast is the distance between the rocket and the spectator changing at that instant?Solution) Distance=s=√(40002+h2), rate of change of distance=ds/dt=(ds/dh)(dh/dt)=(d(√(40002+h2)/dh)(dh/dt),dh/dt=600 ft/sec
Sir,Do you tell me about 5th state of matter i.e Be condensates
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