Answer:
Explanation:
a )
The stored elastic energy of compressed spring
= 1 / 2 k X²
= .5 x 19.6 x (.20)²
= .392 J
b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .
c ) Let h be the distance along the incline which the block ascends.
vertical height attained ( H ) =h sin30
= .5 h
elastic potential energy = gravitational energy
.392 = mg H
.392 = 2 x 9.8 x .5 h
h = .04 m
4 cm .
=
A saw, because wheels and or axles are present in all of the other items. A saw is a wedge, this it does not contain axles or wheels.
Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W × 
/ ( 250 × π × 20 )
1300 × 
= (0.286)² × W × / ( 250 × π × 20 )
solve it and we get W
W = 249.56 × 
so elastic partial width is 2.49 eV
Answer:
0 N
Explanation:
According to Newton's second law, the net force is the mass times the acceleration.
∑F = ma
At constant velocity, the acceleration is 0. So the net force is 0.
