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发表于 2-9-2007 09:58 AM
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intersection 是十字路口，
这题其实要考kinematics 的基本东西， 看你了解如何正确使用
s = ut + 0.5 at^2
v^2 = u^2 + 2as |
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发表于 2-9-2007 09:45 PM
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如果汽车刹车，那么用v^2=u^2+2as 和 s=ut+0.5at^2
v=0 u=20 a=-3.8
那么s=52.6m
不过，当交通灯转红时，用s=ut+0.5at^2,
t=3 u=20 a=-3.8
那么s=42.9m
这个case不懂算不算。
如果继续采油，用s=ut+0.5at^2，
u=20 a=2.3 t=3
那么s=70.35m
车的尾端离十字路口的距离是73.5m，也就是说3.15m的车身从尾端算起还在十字路口里。
十字路口的宽当作是其parallel车移动的方向的长。
不知道对不对？？？ |
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发表于 3-9-2007 12:56 AM
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发表于 3-9-2007 12:19 PM
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我要请教一题有关capacitor,
a rolled paper capacitor is made from strips of metal foil of dimensions 2cm x 40 cmseparated by paper of relative permittivity 2 and thickness 0.002cm .extimate its capacitances.
[ 本帖最后由 yawchoong 于 3-9-2007 12:24 PM 编辑 ] |
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发表于 3-9-2007 03:28 PM
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原帖由 yawchoong 于 3-9-2007 12:19 PM 发表 
我要请教一题有关capacitor,
a rolled paper capacitor is made from strips of metal foil of dimensions 2cm x 40 cmseparated by paper of relative permittivity 2 and thickness 0.002cm .extimate its ...
Let C=capacitance
e= relative permittivity
A=Area of separation capacitor
d=thickness of capacitor
C=eA/d
C=2*[(2*40)x10^-4]/(0.002x10^-2)
C=800C
剑侠, lyt87 不知道对不对？？？
[ 本帖最后由 zxteh 于 3-9-2007 03:30 PM 编辑 ] |
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发表于 5-9-2007 02:23 PM
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hm..你是不是漏了e，permitivity 8.85^-12 ?
即使乘了e,答案还是错。。
书上的答案是2*(Er.Eo.A/d) |
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发表于 5-9-2007 08:43 PM
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原帖由 yawchoong 于 5-9-2007 02:23 PM 发表
hm..你是不是漏了e，permitivity 8.85^-12 ?
即使乘了e,答案还是错。。
书上的答案是2*(Er.Eo.A/d)
soly ,i consider wrongly...
E=ErEo
therefore,
C=2EA/d
C={2*[8.85*10^-12][(2*40)x10^-4]}/(0.002x10^-2)
C=7.08*10^-9C
i know is already wrong...
why capacitance like this too small,how can it function...? |
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发表于 7-9-2007 01:10 PM
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我是新来的…请多指教
An object at rest on the moon surface has a mass of 2.0 kg and weight 3.2 N
An upward force 48 N acts on the object until it reaches a height of 0.50 m above the surface. The object is then released.
What is the maximum height that the object will reach? |
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发表于 7-9-2007 10:46 PM
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对不起，最近比较忙。
g on moon= 3.2 / 2
=1.6 m/s^2
先找velocity once the force is removed,
F = ma, m=2, F=44.8
thus, a = 22.4 m/s^2
v^2 = u^2 + 2as
u=0 a =22.4, s = 0.5
v=4.7 m/s
接下来，用
v^2 = u^2 + 2 a s2 ( s2 是force 离开后的displacement)
这时， u是 刚才的v
a 是 g = -1.6m/s^2 ( -ve 因为往下）
当达到maximum height 时，
v= 0，
所以， s2 = 7 m
s1 = 0.5m
thus, maximum height, h = s1 + s2 = 7.5m
[ 本帖最后由 lyt87 于 11-9-2007 12:58 AM 编辑 ] |
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发表于 10-9-2007 05:17 PM
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wrong....th resultant f acting on the object during the first 0.5 m is 48 N - 3.2 n =44.8 n
upward acceleration =F/m = 44.8 N / 2.o kg =22.4 m s-2 |
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发表于 12-9-2007 07:52 PM
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由于最近剑侠在＂暗拼＂，所以吩咐我来问以下这一题。。。
WHAT IS THE DIFFERENT BETWEEN INTERFERENCE（LIGHT WAVE） AND STATIONARY WAVE？？？
大家帮帮忙。。 |
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发表于 13-9-2007 12:58 PM
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所谓的light wave是propagation waves吧？？？ |
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发表于 1-11-2007 06:37 PM
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请教小弟以下问题：
1。After an U ( A:238, Z: 92) nucleus capture a neutron, it decays in succession a beta-particle followed by another beta-particle. the nuclide produced is
A. Np ( A:240, Z:91 )
B. Pa ( A:240, Z:94 )
C. Pu ( A:238, Z:94 )
D. Th ( A:239, Z:90 )
2. Calculate the total energy released by 1.0g of U-235 which undergoes this nuclear fission. Neglect the mass of electrons produced.
U + n ---> Mo + La + 2n + 7 negative beta-particle
Mass:
U( A: 235, Z: 92)=235.044u
Mo( A: 95, Z: 42)=94.906u
La(A: 139, Z: 57)=138.906u
n(A:1 , Z:0)=1.009u
1u = 931 MeV
多谢大家。。。 |
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发表于 2-11-2007 12:58 AM
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回复 #53 huhuxboy的帖子
question 1:
U ----> Pu + 2e-
i thinks the answer is C.
question 2:
mass defect=m(i)-m(f) i:initial;f:final
=(235.044u+1.009u)-(94.906u+138.906u+2*1.009u)
=0.223u
E=mc^2
=0.223u*[(3.0*10^8)^2]
=3.33810^-11 J /1.60*10^-19
=208*10^6 eV
=208 MeV
or
can juz substitute 1u=931MeV
1u=931MeV
0.223u=----->0.223/1*931
=208MeV |
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发表于 2-11-2007 11:06 PM
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嗯light wave应该是propagation wave 但是我还不知道其答案呢！！！
有谁可以帮忙解答吗？？？？ |
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楼主 |
发表于 3-11-2007 02:44 PM
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其实答案参考书里都有啊..
STATIONARY WAVE DOES NOT TRANSFER ENERGY, THE ADJACSENT AMPLITUDE IS NOT IN PHASE, ... |
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发表于 3-11-2007 05:34 PM
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原帖由 zxteh 于 2-11-2007 12:58 AM 发表 
回复 #53 huhuxboy的帖子
question 1:
U ----> Pu + 2e-
i thinks the answer is C.
question 2:
mass defect=m(i)-m(f) i:initial;f:final
=(235.044u+1.009u)-(94.906u+138 ...
for Q1, 它不是把neutron captured 了吗？为何nucleon number 没有加一呢？？？
for Q2, how abt the energy released per nucleon??? |
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发表于 3-11-2007 05:37 PM
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原帖由 sylvia_r 于 3-11-2007 02:44 PM 发表
其实答案参考书里都有啊..
STATIONARY WAVE DOES NOT TRANSFER ENERGY, THE ADJACSENT AMPLITUDE IS NOT IN PHASE, ...
什么意思是ADJACENT AMPLITUDE IS NOT IN PHASE？？？ |
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发表于 3-11-2007 05:41 PM
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楼主 |
发表于 4-11-2007 07:49 AM
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回复 #58 huhuxboy 的帖子
排写,是我弄错了..
STAIONARY WAVE,
1.WAVE PROFILE DOES NOT MOVES THROUGH THE MEDIUMS..就是它没有在向前走
2.PARICLES BETWEEN TWO CONSECUTIVR NODES VIBRATE WITH DIFFERENT AMPLITUDE
3.PARTICLE BETWEEN TWO NEIGHBOURING NODES VEBRATE IN THE SAME PHASE..就是它们都是一起振动,一起上上下下咯,虽然它们的AMPLITUDE都不一样
4.PARTICLES AT NODES DO NOT VIBRATE AT ALL
5.PRODUCED BY THE SUPERPOSITION OF TWO WAVES MOVING IN OPPOSITE DIRECTIONS
6.DOES NOT TRANSMIT ENERGY |
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