Water vapor enters an insulated nozzle operating at steady state with a velocity of 100 m/s and specific enthalpy of 3445.3 kJ/kg, and exits with specific enthalpy of 3051.1 kJ/kg. The velocity at the exit is most closely

(a) 104 m/s
(b) 636 m/s
(c) 888 m/s
(d) 894 m/s

(a) 104 m/s

Explanation:

enters with specific enthalpy  'h1' = 3445.3 kJ/kg

exits with specific enthalpy 'h2' 3051.1 kJ/kg

velocity 'V1'= 100 m/s

In order to find velocity at the exit 'V2', we use this relation

0 = Qcv - Wcv + m[(h1-h2) +  + g(z1-z2)]

0= [Qcv + m[(h1-h2) + ]

0=( h1-h2) + ( )

0= (3445.3-3051.1)+

0= 394.2 +

-394.2=

-788.4 = 10000-V2²

V2²= 10000+ 788.4

V2²= 10784.44

Taking square root on both sides

V2=103.84

V2≈ 104m/s

Therefore, The velocity at the exit is most closely  104m/s

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

pls mark as brainliest.