Answer: 9.08KW and 16.21KW
Explanation:
The convection over a flat surface with a length of 10 m and a width of 6m.
The mean temperature is (5oC + 12oC)/2 = 8.5oC.
Then find the following properties of air at this temperature from Table A-15:
k = 0.02428 W/m(oC, v= 1.413x10-5 m2/s, and Pr = 0.7340.
find the Reynolds number. Re= VL/v
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This means that the flow becomes turbulent over the plate and we can use the Nusselt number equation for combined laminar and turbulent flow.
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We then use this Nusselt number to find the heat transfer coefficient and the heat transfer.
Check the screen shot for the calculation
Ans
9.08 kW
Then if the wind velocity were doubled, the Re number would be doubled and we would repeat the calculations above, starting with this revised Reynolds number..
Ans
16.21 kW
Answer:
To both observers, the land opposite them is moving to the right.
Explanation:
I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .
https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/
Answer:
A) i) E =α/ [2πrL(εo)]
ii) E=0
iii) E = α/(πrεo)
The graph between E and r for the 3 cases is attached to this answer ;
B) i) charge on the inner surface per unit length = - α
ii) charge per unit length on the outer surface = 2α
Explanation:
A) i) For r < a, the charge is in the cavity and takes a shape of the cylinder. Thus, applying gauss law;
EA = Q(cavity) / εo
Now, Qcavity = αL
So, E(2πrL) = αL/εo
Making E the subject of the formula, we have;
E =α/ [2πrL(εo)]
ii) For a < r < b; since the distance will be in the bulk of the conductor, therefore, inside the conductor, the electric field will be zero.
So, E=0
iii) For r > b; the total enclosed charge in the system is the difference between the net charge and the charge in the inner surface of the cylinder.
Thus, Qencl = Qnet - Qinner
Qinner will be the negative of Qnet because it should be in the opposite charge of the cavity in order for the electric field to be zero. Thus;
Qencl = αL - (-αL) = 2αL
Thus, applying gauss law;
EA = Qencl / εo
Thus, E = Qencl / Aεo
E = 2αL/Aεo
Since A = 2πrL,
E = 2αL/2πrLεo = α/(πrεo)
B) i) The charge on the cavity wall must be the opposite of the point charge. Therefore, the charge per unit length in the inner surface of the tube will be - α
ii)Net charge per length for tube is +α and there is a charge of - α on the inner surface. Thus charge per unit length on the outer surface will be = +α - (- α) = 2α
Answer:
34.17°C
Explanation:
Given:
mass of metal block = 125 g
initial temperature = 93.2°C
We know
..................(1)
Q= Quantity of heat
m = mass of the substance
c = specific heat capacity
c = 4.19 for H₂O in
= change in temperature
Now
The heat lost by metal = The heat gained by the metal
Heat lost by metal =
Heat gained by the water =
thus, we have
=
⇒
Therefore, the final temperature will be = 34.17°C