A golf club strikes a 0.037-kg golf ball in order to launch it from the tee.?
2016-06-14 7:53 am
assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 6900 N, and is in contact with the ball for a distance of 0.0080 m. With what speed does the ball leave the club?
回答 (2)
2016-06-14 8:07 am
F = m a
6900 = 0.037 a
a = 6900/0.037 m/s²
v² = vₒ² + 2as
v² = 0² + 2 × (6900/0.037) × 0.0080
v = √[2 × (6900/0.037) × 0.0080] m/s
v = 54.6 m/s
Speed that the ball leaves the club = 54.6 m/s
6900 = 0.037 a
a = 6900/0.037 m/s²
v² = vₒ² + 2as
v² = 0² + 2 × (6900/0.037) × 0.0080
v = √[2 × (6900/0.037) × 0.0080] m/s
v = 54.6 m/s
Speed that the ball leaves the club = 54.6 m/s
2016-06-15 3:42 pm
acceleration a = F/m = 6.9*10^6/37 = 186,486 m/sec^2
V = √2*a*d = √2*186,486*0,0080 = 54.624 m/sec
V = √2*a*d = √2*186,486*0,0080 = 54.624 m/sec
收錄日期: 2021-04-18 15:04:23
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https://hk.answers.yahoo.com/question/index?qid=20160613235345AAEkJg1