Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is 
, and the steam properties as, A-4E:
Using the energy balance for the system:
Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:
Entropy generation in the heat exchanger could be defined using the entropy balance on the system:
Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R
I think the answer is c. but I think it depends on how many zebras you have
I would tell him, in the kindest, most gentle way I could manage,
to fahgeddaboudit.
The total amount of energy doesn't change. Energy is never created,
and it never disappears. If you have some energy, then it had to come
from somewhere, and if you used some energy, then it had to go
somewhere.
You can never get more energy out of the electromotor than you put into it,
and in the real world, you can't even get THAT much out, because some
of it is always used on the way through.
Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
"Free Energy" on the internet, relax, and enjoy the show. They are all
fakes. They may not all be intentionally meant to fool you, but they are
all impossible.
Answer:
0.0025116weber/m²
Explanation:
Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).
Mathematically;
B = ¶/A
¶ = BA
Given B = 0.23Tesla which is the magnitude of the magnetic field
Dimension of the rectangular loop = 7.8 cm by 14 cm
Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm
= 109.2cm²
Converting this value to m²
Area of the loop = 109.2 × 10^-4
Area of the loop = 0.01092m²
Magneto flux = 0.23×0.01092
Magnetic flux = 0.0025116weber/m²
For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as
F= 618.9 N
<h3>What is the centripetal
force?</h3>
Generally, the equation for the angular speed is mathematically given as
w = v/R
Therefore
w= 4.7/1.8
w= 2.611 rad/s
Where total momentum
Tm= 642.96 + 272.32
Tm= 915.28
and total inertia
Ti= 184 + 246.24
Ti= 430.24
In conclusion, centripetal force
F= mrw^2
F = m*R*w2^2
F = 76*1.8*2.127^2
F= 618.9 N
Read more about mass
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CQ
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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?
