If a firefighter slides down a pole with an acceleration of 7.3 m/s^2 ...
If he starts at rest and slides for a distance of 3.6 m, what is the firefighter's final speed?
Thanks in advance and have a great day
Simple Physics Question?
2015-11-02 3:49 am
回答 (2)
2015-11-02 8:32 am
V = √2ad = √7.3*2*3.6 = 7.250 m/sec
2015-11-02 5:10 am
(g - 7.3) = (9.8 - 7.3) = 2.5m/sec^2 acceleration.
V = sqrt.(2ah) = 4.24m/sec.
V = sqrt.(2ah) = 4.24m/sec.
收錄日期: 2021-04-21 14:50:20
原文連結 [永久失效]:
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