The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
1 answer:
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Answer:
W(r,out) = 5.81 MW
= 86.1 %
Explanation:
we use here steam table for get value of h1, s1 etc
so use for 6MPa and 600 degree
Enthalphy of steam h1 = 3658.8 kJ/kg
Entropy of steam s 1 is = 7.1693 kJ /kg.K
and
for 50 kPa and 100 degree
Enthalphy of steam h2 = 2682.4 kJ/kg
Entropy of steam s2 is = 7.6953 kJ /kg.K
so we use here energy balance equation that is
..............1
put here value and we get m
m =
solve it we get
m = 5.156 kg/s
so by energy balance equation
m1 = m
2 + W(r,out)
W(r,out) = m(1 -
2)
W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)
W(r,out) =
W(r,out) = W(a,out) - m.To.(s1-s2) ........................2
put here value
W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)
W(r,out) = 5908.19 = 5.81 MW
and
second law deficiency is
=
..............................3
put here value
= \frac{5}{5.81}
= 86.1 %
The resultant of the two vectors is 141 km/h.
Explanation:
When two vectors are at right angles to each other, their resultant can be calculated by using the Pythagorean's theorem:
where
A is the magnitude of the first vector
B is the magnitude of the second vector
In this problem, we have two velocity vectors of magnitude
A = 100 km/h
B = 100 km/h
Substituting into the formula, we find their resultant:
Learn more about vector addition here:
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