| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 13:18:55 IST
|
|
|
if d earth were to complt a day in 1 hr,,,den wat wud be d wt of man as registrd on a weighin machine???????
|
|
|
|
22 Mar 2008 13:24:53 IST
|
|
|
On the equator,
the weight recorded would be minimum
this is due to centrifugal force provided by rotation of earth
F=centrifugal force
F=mw2r
w=2pi/(1 hour)
=2 pi/(3600 sec)
a=w2r=(3*10-6)6400000
a=w2r=19.49 m/s2since this is greater than g, the man would be thrown off and feel weightless
|
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:29:44 IST
|
|
|
bt u hav to find a general formula fr dat,,,,,,,,its nt askin abt d equator........
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:33:36 IST
|
|
|
If he is at a latitude of
A degrees,
then centrifugal force opposing gravity=mw2(rcosA)cosA=mw2rcos2A
since radius becomes rcosA and horizontal component=acosA
so his weight as general expression would be
m(g-w2rcos2A)
value is
m(9.8-19.5cos2A)
|
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
this reply: 14 points
(with 2
in 4 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:36:04 IST
|
|
|
akhil..very nice answer..gr8 work
|
Nitwit Blubber Odment Tweak
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:40:41 IST
|
|
|
can u show d diagram????????
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:41:52 IST
|
|
|
click and zoom
|
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:49:46 IST
|
|
|
wats a hre??
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 13:52:04 IST
|
|
|
sorry?
hre??
which part is it in?
|
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
22 Mar 2008 23:12:12 IST
|
|
|
sorry
a is centrifugal acceleration!
|
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Mar 2008 22:56:09 IST
|
|
|
Q) if the earth were to complete a day in 1 hr then wat wud be the weight of man as registrd on a weighing machine? Sol) Effect of rotation of earth on weight mg of an object is given by Efective weight = W' = m(g-w2rcos2A) where W = mg = weight in the absence of rotation w = angular speed of the earth A = Latitude of earth at the point where object is kept A = 0 deg, at equator (thus effect od rotationis maximum on equator) = 90 deg, at poles, (thus effect is nil at the poles) Now if earth completes its rotation in 1hr, then itsangular speed is increased by 24 times, thus w' = 24w So the weight mg of the man will be W'' = m(g- 576w2rcos2A) Thus W'' = m(g-w2rcos2A) - 575mw2rcos2A or W'' = W' - 575mw2rcos2A Thus teh weight of man is reduced by a 575w2rcos2A on weighing machine
|
The most incomprehensible thing about the world is that it is
at all comprehensible. |
this reply: 10 points
(with 2
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
![[Thumb - earth.JPG]](/upload/2008/3/22/c49606eb603c0a478d370284696595c1_37348.JPG_thumb)
