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
The variation of water depth at spreading centers (ridges) controlled by isostasy as convective cooling cools the rocks much more effectively the than heat conduction.
<h3>What is convective heat transfer?</h3>
When heat transfer takes place between the two fluids in direct or indirect contact.
The lithosphere cools when it moves away from the ridge axis by sea floor spreading. The cooler rocks have low density, so the sea floor gets deeper as the lithosphere gets more dense.
Thus, the convective cooling cools the rocks much more effectively the than heat conduction.
Learn more about convective heat transfer
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Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As

For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Answer:
<em>The magnitude of the magnetic field will act in a direction towards me.</em>
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Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. <em>The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field.</em> In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
The answer is 294j by puting it in kinetic energy formula.