Term-1 physic讨论区

weston123453

weston123453 发表于 11-6-2014 05:56 PM
接下来的那题不明白它的意思勒。。。。。highest point of motion not means the velocity is equal t ...

oo highest point is vertical component of velocity=0, horizontal still the same
horizontal component of velocity at highest point is  v cos 60= 0.5 v
initail K.E is K
at highest point 就是 1/2 m(0.5v)^2=(1/2mv)/4
                                                    =1/4 K

doeramon

我只是b)不会做罢了
A missile is fired with a speed of 500ms^-1 from the ground at angle 45; to the horizontal
a)calculate maximum range ,R
ans: 2.55x10^4 m

b) What is the maximum height above the ground when its range is maximum?
ans: 6.37x10^3 m

谁有pelangi的书？我要chapter 2 quick check 5 objective的答案

doeramon

weston123453 发表于 11-6-2014 07:52 PM
oo highest point is vertical component of velocity=0, horizontal still the same
horizontal compon ...

1/2 m(0.5v)^2=(1/2mv)/4
                        =1/4 K

看不懂你怎样做成1/4 k
(1/2mv)/4是 1 over 2 乘mv 然后再除4还是1 over 2 mv 然后再除4?

weston123453

doeramon 发表于 11-6-2014 11:52 PM
1/2 m(0.5v)^2=(1/2mv)/4
                        =1/4 K

有分别吗？
(1/2mv)/4=k/4

weston123453

doeramon 发表于 11-6-2014 11:04 PM
我只是b)不会做罢了
A missile is fired with a speed of 500ms^-1 from the ground at angle 45; to the  ...

用FORMULA==
H=(u^2sin^2 theta)/2g

doeramon

weston123453 发表于 12-6-2014 01:14 AM
有分别吗？
(1/2mv)/4=k/4

0.5mv^2为什么最后不见了

doeramon

weston123453 发表于 12-6-2014 01:16 AM
用FORMULA==
H=(u^2sin^2 theta)/2g

我被max range搞乱了

八卦一下，你的老师叫你背formula?马来老师吗？

weston123453

doeramon 发表于 12-6-2014 11:18 AM
我被max range搞乱了

八卦一下，你的老师叫你背formula?马来老师吗？

我的老师是华人，还没教到第2课。。。
formula我没有背，用v^2=u^2+2as就可以变出来了。。。

weston123453

doeramon 发表于 12-6-2014 10:56 AM
0.5mv^2为什么最后不见了

懂了？
你右边那个是错的，不是1/（2mv^2).........是1/2(mv^2)

MichelleLow

Hello~谁能帮我解决这问题吗？

A force of 6.00N acts at the origin in a direction of 30.0degree above the position x-axis. A second force of 5.00N acts at the origin in the direction of the positive y-axis, Find the magnitude and direction of the resultant force.

HenryTeo66

我发现好少人拿physics 哦， 我的学校才5个人拿physics...顺便想问下你们有用微信吗， 如果有可以来加我吗id 是 henryteo96, 我希望大家可以时常分享好的资料还有讨论些问题之类的。。。

YetSin

MichelleLow 发表于 2-7-2014 03:25 PM
Hello~谁能帮我解决这问题吗？

A force of 6.00N acts at the origin in a direction of 30.0degree abo ...

F_x=6 cos 30
F_y=5+6sin30
F_resultant=\sqrt[(5+6sin30)^2+(6cos30)^2]
Direction = arctan [(5+6sin30)/6cos30] (anticlockwise from positive x-axis)

MichelleLow

YetSin 发表于 2-7-2014 04:42 PM
F_x=6 cos 30
F_y=5+6sin30
F_resultant=\sqrt[(5+6sin30)^2+(6cos30)^2]

i got it, thank you so much

MichelleLow

HenryTeo66 发表于 2-7-2014 03:43 PM
我发现好少人拿physics 哦， 我的学校才5个人拿physics...顺便想问下你们有用微信吗， 如果有可以来加我吗i ...

me too, 我的学校才8位拿而已

MichelleLow

我有个疑问在projectile那part, an object fall from a certain height,ex.= 5m,its acceleration is +, why the displacement also +??not -ve?

YetSin

MichelleLow 发表于 2-7-2014 06:02 PM
我有个疑问在projectile那part, an object fall from a certain height,ex.= 5m,its acceleration is +, wh ...

it all depends on your sign convention, the positive direction of displacement, velocity and accelerationa are always the same. If you define downward as your positive direction, then everything downwards will be considered as positive and upward will be negative.

In your question, since the question only involves falling, define downward as positive will be more convenient. Of course, you can still define downward as negative as usual and still get the same result.

weston123453

各位的老师教到哪里了？觉得自己快跟不上了。。。目前我的老师教到第4课

weston123453

weston123453 发表于 4-7-2014 10:34 AM
各位的老师教到哪里了？觉得自己快跟不上了。。。目前我的老师教到第4课

是教完了第4课==

树熊123

weston123453 发表于 4-7-2014 10:38 AM
是教完了第4课==

我们的老师还在第三课@@

树熊123

HenryTeo66 发表于 2-7-2014 03:43 PM
我发现好少人拿physics 哦， 我的学校才5个人拿physics...顺便想问下你们有用微信吗， 如果有可以来加我吗i ...

我们学校才六个