The rubber cord of a catapult has a cross sectional area 1mm^2
and total unstretched length
10cm. It is stretched to 12cm and then released to project a missile of mass 5gm. Taking 'Y'
for rubber 5x10^8
N/m2
, the velocity of projection is
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Answer:
The rubber cord of a catapult has a cross-sectional area 1mm2 and total unstretched length 10cm. It is stretched to 12cm and then released to project a missile of mass 5gm.
Explanation:
F=
l
YAΔl
=
10×10
−
2
5×10
8
×10
−
6×2×10
−
2
energy=
2
1
×F×Δl=
2
1
×
10×10
−
2
5×10
8
×10
−
6×2×10
−
2
×2×10
−
2=
2
1
mv
2
So,
10×10
−
2×5×10
−
3
5×10
8
×10
−
6×2×10
−
2×2×10
−
2
=v
2
v=20m/s
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