A force of 40 N acting on abody of mass 10 Kg changes its velocity from 5m/s to 25m/s determime acceleration and time for which force acts anddistance answered
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Answered by
0
f=m×a
40=10×a
a=40/10=4m/s2
a=v-u/t
4=25-5/t
t=20/4=5sec
s=ut+1/2at2
s=5×5+1/2×4×5×5
s=25+50
s=75m
if this helped u i will be thankful
40=10×a
a=40/10=4m/s2
a=v-u/t
4=25-5/t
t=20/4=5sec
s=ut+1/2at2
s=5×5+1/2×4×5×5
s=25+50
s=75m
if this helped u i will be thankful
Answered by
2
S=1/2(u+v)t
=>ut+1/2at²
=(5*0.2)+1/2*4*(0.2)²
=1+(2*0.04)
=1+0.08
=1.08m
=>ut+1/2at²
=(5*0.2)+1/2*4*(0.2)²
=1+(2*0.04)
=1+0.08
=1.08m
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ShreyaBhowmick:
Oh u have already kept my nickname....!
ya
hmm...
where r u
Excuse me..
oh soryyyyyy
hey ur last step is wrong
20/t=4
then 20/4=t
that is 5
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