A bullet having a mass of 0.005 kg is moving with a speed of 100m/s . It penetrates into a bag of sand and its brought to rest after moving 25 cm into a bag .Find the decelerating force on the bullet .Also calculate the time in which it is brought to rest ???
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retardation is 100 and time is 1/200 seconds
xheikhfiza:
but formula is s=vit 1/2 at^2
so ut ????
25/100 = 100t -1/2 at^2
0= 100 -at
as vis 0
at = 100
now 1=400t-200t
1 = 200t
I dont understand
what for u uz
Answered by
4
hiii dear_______
"good morning" __________________
here is your answer_______________
___________
_______________
It is given that ;
S = 25cm = 0.5m
S =( Vo)(ts) - {(a)(ts)^2 / 2}
where (ts) denotes the time for which the object moved until stop .
also for velocity ,
v = Vo - a(ts)
and for stop Timing ;
0 = Vo - a(ts)
(ts) = Vo / a
plugging this formulla into formulla for S Obtain :
(ts) = 2S/ Vo = 0.005 sec.
hence the acceleration is
|a| = Vo / (ts) = {100m/s} / {0.005 sec.}
|a| = 20000 m/sec 2
(actually this acceleration is negative , but one needs the absolute value only)
hence ,
|F| = m|a|
|F| = 20000 * 0.005
= 100 N
"good morning" __________________
here is your answer_______________
___________
_______________
It is given that ;
S = 25cm = 0.5m
S =( Vo)(ts) - {(a)(ts)^2 / 2}
where (ts) denotes the time for which the object moved until stop .
also for velocity ,
v = Vo - a(ts)
and for stop Timing ;
0 = Vo - a(ts)
(ts) = Vo / a
plugging this formulla into formulla for S Obtain :
(ts) = 2S/ Vo = 0.005 sec.
hence the acceleration is
|a| = Vo / (ts) = {100m/s} / {0.005 sec.}
|a| = 20000 m/sec 2
(actually this acceleration is negative , but one needs the absolute value only)
hence ,
|F| = m|a|
|F| = 20000 * 0.005
= 100 N
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