a force is applied on a body of mass 0.9kg that is at rest.the force is applied for a duration of 5 sec and as a result the body covers a distance of 250m.what is the force?
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m=0.9 kg
F=ma so a=F/m=F/0.9
Acoording to 3rd law of linear motion
s=ut+1/2*at²
since u=0 and a=F/0.9
250=1/2*F/0.9*25
10=1/2*F/0.9
F=9*2=18 N
PLEASE MARK AS BRAINLIEST IF HELPFUL!!!
F=ma so a=F/m=F/0.9
Acoording to 3rd law of linear motion
s=ut+1/2*at²
since u=0 and a=F/0.9
250=1/2*F/0.9*25
10=1/2*F/0.9
F=9*2=18 N
PLEASE MARK AS BRAINLIEST IF HELPFUL!!!
pinkspink23:
i am nt getting the right answer
sorry for inconvenience
have corrected the answer
did u get 18N?
yes presh refresh on ur screen
u will see the edited answer
ohhh yea....tysm !!!
plz mark as brainliest
done
could u see the next question i posted
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