For the steady flow process, the first law is written like
DH + Du2/2 + gDz = Q + Ws
since there is no shaft work, Ws = 0
and flow is horizontal, Dz = 0
Therefore,
DH + Du2/2 = Q
substituting for the quantities,
(2726.5 - 334.9) x 1000 + (200^2 - 3^2)/2 = Q (in terms of J/kg)
Q = 2411.1 kJ/kg
Heat transferred through the coil per unit mass of water = 2411.1 kJ
Answer:
the potentail of kinetic and potential energy
Explanation:
first explain the concept of kinetic energy (what it is and what its used for) and give examples (cars, a basketball thrown across a hall, and airplane), and do the same with potential energy (the energy an object stores, example: a streched rubber band)
Answer:
Explanation:
We shall apply conservation of mechanical energy .
initial kinetic energy = 1/2 m v²
= .5 x m x 12 x 12
= 72 m
This energy will be spent to store potential energy . if h be the height attained
potential energy = mgh , h is vertical height attined by block
= mg l sin20 where l is length up the inclined plane
for conservation of mechanical energy
initial kinetic energy = potential energy
72 m = mg l sin20
l = 72 / g sin20
= 21.5 m
deceleration on inclined plane = g sin20
= 3.35 m /s²
v = u - at
t = v - u / a
= (12 - 0) / 3.35
= 3.58 s
it will take the same time to come back . total time taken to reach original point = 2 x 3.58
= 7.16 s
There you go!
as it is unnecessary to find velocity, so use the equation with v
remember to take all directions downwards as it is more convenient to calculate
after substituting the values, you have to reject the negative answed as time must be+
:)
The boiling pot of water is completely open
