Please help with physics? Thank you in advance!?
2018-01-18 7:42 am
A 1380 kg car starts from rest at the top of a 28.0 m long hill inclined at 11.0°. Ignoring friction, how fast is it going when it reaches the bottom of the hill?
回答 (2)
2018-01-18 8:00 am
✔ 最佳答案
Take g = 9.81 m/s² F = ma
mg sin11.0° = ma
a = g sin110°
v² = u² + 2as
v² = u² + 2gs sin110°
v² = 0² + 2(9.81)(28.0) sin11.0°
v = √[2(9.81)(28.0) sin11.0°]
v = 10.2 m/s
Speed when the car reaches the bottom of the hill = 10.2 m/s
2018-01-18 7:52 am
the height of the hill is 28 sin 11 = 5.34 m
PE = KE
mgh = ½mV²
V = √(2gh) = √(2•9.8•5.34) = 10.2 m/s
PE = KE
mgh = ½mV²
V = √(2gh) = √(2•9.8•5.34) = 10.2 m/s
收錄日期: 2021-04-18 18:14:31
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