Answer:
Explanation:
We shall apply law of conservation of momentum in space to know the velocity of combination after the impact
m₁v₁ = m₂v₂
.1 x 4 = ( 1 + .1 ) v₂
v₂ = .3636 m /s
1 )
Kinetic energy of the combination
= 1/2 x 1.1 x ( .3636)²
= 7.3 x 10⁻² J
2 )
Initial kinetic energy of the system
= 1/2 x 0.1 x 4²
= 0.8 J
Final kinetic energy of the system = 7.3 x 10⁻²
Loss of energy = .8 - .073
= .727 J
This energy was converted into internal energy of the system .
3 )
increase in entropy = dQ / T
Here dQ = .727 J
T = 300 ( Constant )
dQ / T = 2.42 X 10⁻³ J/K
Answer:
120s^-1
Explanation:
v=12v
I=10A
and since rate is with time, therefore rate=energy/time.
H=IV
10×12=120/s
therefore the rate is 120s^-1
Answer:Direct them straight across
Explanation:
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
