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发表于 30-7-2010 08:50 PM
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回复 340# vick5821
Torque = Ia
s= ut + at^2 / 2 (用angular form的symbol, 这里打不出)
work done = Torque x angular displacement
power deliver to the wheel in 60s = 257 kW |
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发表于 30-7-2010 09:56 PM
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The answer is not this wor.. |
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发表于 30-7-2010 10:13 PM
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回复 342# vick5821
那答案是什么? 下次可以连答案一起放吗 |
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发表于 30-7-2010 10:20 PM
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发表于 30-7-2010 11:08 PM
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回复 344# vick5821
Torque = Ia
v= u + at(用angular form的symbol, 这里打不出)
work done
= Torque x angular displacement
= Torque x vt
power deliver
= Work done / 60
= 8571 W |
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发表于 31-7-2010 11:12 AM
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发表于 31-7-2010 12:20 PM
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回复 346# vick5821
vt = s
angular velocity is assumed to be constant |
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发表于 31-7-2010 12:32 PM
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In this case, angular velocity is not constant wor..changing with time leh |
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发表于 31-7-2010 12:32 PM
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发表于 31-7-2010 01:56 PM
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回复 349# vick5821
写错了...是constant angular acceleration |
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发表于 31-7-2010 07:55 PM
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then s=vt is for constant angular velocity de leh.. |
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发表于 31-7-2010 08:42 PM
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URGENT!!!
A cyclist is riding with a speed of 36km/hr. As he approaches a circular turn on the road of radius 50m, he applies brakes and reduces his speed at constant rate of 0.5m/s^2. What is the magnitude and direction of net acceleration of cyclist on circular road? |
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发表于 31-7-2010 08:48 PM
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发表于 31-7-2010 08:57 PM
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2.1 ms-2?????
vick5821 发表于 31-7-2010 08:48 PM 
yaya
我刚刚算到答案了
不过还是谢谢你
哈哈 |
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发表于 31-7-2010 09:10 PM
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发表于 31-7-2010 09:43 PM
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To #352 and #355
Firstly, apply the formula a_c = v^2/r = (10)^2/50 = 2ms^-2, since 36km/h = 10ms^-1, and the direction of this acceleration a_c is "to the center".
Then, draw the diagram to determine the resultant acceleration. Since the deceleration is 0.5ms^-2, the direction is to the back of the cyclist.
Apply the pythagoras' theorem and determine the magnitude of resultant acceleration: a_r = sqrt (2^2 + 0.5^2) = 2.06 ms^-2 (3 sf)
Apply the formula tangent theta = sin theta / cos theta = 0.5/2 and determine the direction of the resultant force: theta = tan^-1 (3/4) = 36.87 degree. Since it is lying in the fourth quadrant, theta_r = -36.87 degree. |
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发表于 31-7-2010 10:10 PM
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To #352 and #355
Firstly, apply the formula a_c = v^2/r = (10)^2/50 = 2ms^-2, since 36km/h = 10ms ...
数学系 发表于 31-7-2010 09:43 PM
thanks a lot... |
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发表于 31-7-2010 10:18 PM
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回复 357# Lov瑜瑜4ever
不用客气。其实你会做那题，不用道谢的 
今年upper six? |
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发表于 31-7-2010 10:47 PM
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I dun understand the part to find the angle..!! argh.can explain?? trhx |
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发表于 31-7-2010 11:05 PM
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回复 359# vick5821
Well, first you have to know the direction of the centripetal force; it is toward the center.
Second, you have to know the direction of the deceleration; it must be opposite to the direction of motion.
Finally, draw a diagram to determine the direction of resultant force. |
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