Answer:
The angular velocity is 7.56 rad/s
the maximum water height is 2 ft
Explanation:
The z-position as a function of r is equal to
(eq. 1)
where
h0 = initial height = 1 ft
w = angular velocity
R = radius of the cylinder = 1.5 ft
zs(r) = 0 when the free surface is lowest at the centre
Replacing and clearing w
if you consider the equation 1 for the free surface at the edge is equal to
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W =
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
8861.75 m approximately 8862 m
Explanation:
We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.
for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that
For simplicity we work with
despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.
In accord with the formula the "normal" or "standard" weight of an object is given by
when
, so we need to find the value of
that makes
meaning that the original weight decrease by a 0.30%, so now we operate...
now we group like terms on the same sides
we cancel equal tems on both sides and obtain that