URGENT! I only need the last question. 0.7533 ISWRONG. An example is given, however, I keep getting thewrong answer. With explanation pls!
The length of nylon rope from which a mountain climber is suspended has a force constant of 1.55×104 N/m. (a) What is the frequency at which he bounces, given his mass plus equipment to be 81.5 kg ? Hz (b) How much would this rope stretch to break the climber's fall, if he free falls 2.00 m before the rope runs out of slack? m (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used. bounce frequency - Hz distance distance
Inserting knew =k/2, we obtain y2, new =−(k2mg)±[(k2mg)2+2(k2mg)y1]1/2. Inserting values, we have y2,new=−1.25×104 N/m2(76.0 kg)(9.80 m/s2)±[1.25×104 N/m2(76.0 kg)(9.80 m/s2)]2+2[1.25×104 N/m2(76.0 kg)(9.80 m/s2)](2.00 m) or y2, new =−1.10 m
URGENT! I only need the last question. 0.7533 IS WRONG. An example is given, however, I keep getting the wrong answer. W
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answerhappygod
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URGENT! I only need the last question. 0.7533 IS WRONG. An example is given, however, I keep getting the wrong answer. W
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