A 2.4 kg purse is dropped from the top of the leaning tower of Pisa and falls 55m before reaching the ground with a speed of 27 m/s.?
回答 (2)
2018-01-17 1:50 pm
Take g = 9.81 m/s²
Denote f as the average force of air friction.
Initial velocity, u = 0 m/s
Final velocity, v = 27 m/s
displacement, s = 55 m
v² = u² + 2as
27² = 0 + 2a(55)
a = 729/110 m/s² = 6.63 m/s²
F = ma
mg - f = ma
(2.4) (9.81) - f = (2.4)(6.63)
f = (2.4) (9.81) - (2.4)(6.63)
Average force of air friction, f = 7.6 N
Denote f as the average force of air friction.
Initial velocity, u = 0 m/s
Final velocity, v = 27 m/s
displacement, s = 55 m
v² = u² + 2as
27² = 0 + 2a(55)
a = 729/110 m/s² = 6.63 m/s²
F = ma
mg - f = ma
(2.4) (9.81) - f = (2.4)(6.63)
f = (2.4) (9.81) - (2.4)(6.63)
Average force of air friction, f = 7.6 N
2018-01-17 11:42 am
Its acceleration = (v^2/2d) = (27^2/110) = -6.63m/sec^2.
Friction force = (ma) = (2.4 x 6.63) = 15.9N.
Friction force = (ma) = (2.4 x 6.63) = 15.9N.
收錄日期: 2021-04-18 18:03:16
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https://hk.answers.yahoo.com/question/index?qid=20180117031927AA6x56o