Answer:
73.13°
Explanation:
According to snell's law,
n1sinθi = n2sinθr
n1/n2 = sinθr/sinθi
Critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°
This means i=C and r = 90°
The Snell's law formula will become
n1/n2 = sinC/sin90°
n2/n1 = 1/sinC
Where n1 is the refractive index of the less dense medium = 1.473
n2 is the refractive index of the denser medium = 1.540
Substituting the values in the formula,
1.540/1.473 = 1/sinC
1.045 = 1/sinC
SinC = 1/1.045
SinC = 0.957
C = sin^-1(0.957)
C = 73.13°
The answer is a) the hunter
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
We will use the formula:
Q = ml
where Q is the heat, m is the mass of substance, and l is the latent heat of vaporization of the substance.
Q = 0.2 x 2.26 x 10³
Q = 452 kJ
Answer:
There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.
Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.
