Answer:7989.86KW
Explanation:
Given data
=3100 KJ/kg
=10kg/s
=30m/s
=3m
=2300KJ/kg
=45m/s
=0
using steady flow energy equation which is

+
=


+
=
+
-
=
=7989.8673 KW
Answer:
Q=127.66W
L=9.2mm
Explanation:
Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.
To solve any problem related to heat transfer, the general equation is used
Q = delta / R
Where
Q = heat
Delta = the temperature difference
R = is the thermal resistance by conduction, convection and radiation
to solve this problem we propose the previous equation
Q = delta / R
later we find R
![R=[tex]r=\frac{6L1}{AK1} +\frac{5L2}{AK2}+\frac{1}{Ah}](https://tex.z-dn.net/?f=R%3D%5Btex%5Dr%3D%5Cfrac%7B6L1%7D%7BAK1%7D%20%2B%5Cfrac%7B5L2%7D%7BAK2%7D%2B%5Cfrac%7B1%7D%7BAh%7D)

Q=(25-(-5))/0.235=127.66W
part b
we use the same ecuation with Q=127.66
Q = delta / R
Δ
Answer:
It mean to equip more tools.
Explanation:
I hope this helps.
<u>Explanation</u>:
(a)
![f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A](https://tex.z-dn.net/?f=f%3DA%2BA%20B%2BA%20C%5C%5C%3DA%5B1%2BB%2BC%5D%3DA%20%5Cquad%5B1%2Bx%3D1%5D%5C%5CF%3DA)
No gate is required to implement this function
(b)

Note: Refer the first image.
(c)
Note: Refer the second image
(d)
![\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%3D%26%20A%20B%20%5Cbar%7Bc%7D%2B%5Coverline%7BA%2B%5Cbar%7Bc%7D%7D%20%5C%5C%3D%26%20A%20B%20%5Cbar%7Bc%7D%2B%5Cbar%7BA%7D%20%5Cbar%7Bc%7D%3D%5Cbar%7BA%7D%20B%20%5Cbar%7Bc%7D%2B%5Cbar%7BA%7D%20c%20%5C%5Cf%3D%26%20%5Cbar%7BA%7D%5Bc%2BB%20%5Cbar%7Bc%7D%5D%20.%20%5C%5C%26%20f%3D%5Cbar%7BA%7D%20B%2B%5Cbar%7BA%7D%20c%3D%5Cbar%7BA%7D%28B%2Bc%29%5Cend%7Baligned%7D)
Note: Refer the third image
(e)
![\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%3D%26%20A%20%5Cbar%7BB%7D%2B%5Cbar%7BB%7D%20C%2BA%20%5Cbar%7BB%7D%20%5C%5C%26%3D%5Cbar%7BB%7D%5BA%2B%5Cbar%7BA%7D%2Bc%5D%20%5C%5C%26%3D%5Cbar%7BB%7D%5B1%2BC%5D%5Cend%7Baligned%7D)

(f)
![\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Df%20%26%3DA%20B%20C%2BA%20B%20D%2BA%20B%20C%20%5C%5C%26%3DA%20B%5BC%2BC%5D%2BA%20B%20D%20%5C%5C%26%3DA%20B%2BA%20B%20D%20%5C%5C%26%3DB%5BA%2BA%20D%5D%20%5C%5C%26%3DB%5BA%2BD%5D%20%5C%5C%5Ctherefore%20%26%20A%3DB%5BA%2BD%5D%5Cend%7Baligned%7D)
Note: Refer the fourth image
Is this even a question ?