Physics, asked by munibharath8295, 9 months ago

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​

Answers

Answered by SamikBiswa1911
1

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

Similar questions