Answer:
The temperature T= 648.07k
Explanation:
T1=input temperature of the first heat engine =1400k
T=output temperature of the first heat engine and input temperature of the second heat engine= unknown
T3=output temperature of the second heat engine=300k
but carnot efficiency of heat engine =
where Th =temperature at which the heat enters the engine
Tl is the temperature of the environment
since both engines have the same thermal capacities <em>
</em> therefore 
We have now that
multiplying through by T
multiplying through by 300
-
The temperature T= 648.07k
Answer:
Check the explanation
Explanation:
Code
.ORIG x4000
;load index
LD R1, IND
;increment R1
ADD R1, R1, #1
;store it in ind
ST R1, IND
;Loop to fill the remaining array
TEST LD R1, IND
;load 10
LD R2, NUM
;find tw0\'s complement
NOT R2, R2
ADD R2, R2, #1
;(IND-NUM)
ADD R1, R1, R2
;check (IND-NUM)>=0
BRzp GETELEM
;Get array base
LEA R0, ARRAY
;load index
LD R1, IND
;increment index
ADD R0, R0, R1
;store value in array
STR R1, R0,#0
;increment part
INCR
;Increment index
ADD R1, R1, #1
;store it in index
ST R1, IND
;go to test
BR TEST
;get the 6 in R2
;load base address
GETELEM LEA R0, ARRAY
;Set R1=0
AND R1, R1,#0
;Add R1 with 6
ADD R1, R1, #6
;Get the address
ADD R0, R0, R1
;Load the 6th element into R2
LDR R2, R0,#0
;Display array contents
PRINT
;set R1 = 0
AND R1, R1, #0
;Loop
;Get index
TOP ST R1, IND
;Load num
LD R3,NUM
;Find 2\'s complement
NOT R3, R3
ADD R3, R3,#1
;Find (IND-NUM)
ADD R1, R1,R3
;repeat until (IND-NUM)>=0
BRzp DONE
;load array address
LEA R0, ARRAY
;load index
LD R1, IND
;find address
ADD R3, R0, R1
;load value
LDR R1, R3,#0
;load 0x0030
LD R3, HEX
;convert value to hexadecimal
ADD R0, R1, R3
;display number
OUT
;GEt index
LD R1, IND
;increment index
ADD R1, R1, #1
;go to top
BR TOP
;stop
DONE HALT
;declaring variables
;set limit
NUM .FILL 10
;create array
ARRAY .BLKW 10 #0
;variable for index
IND .FILL 0
;hexadecimal value
HEX .FILL x0030
;stop
.END
Answer:
Define Variables and Use List methods to do the following
Explanation:
#<em>Conjoins two lists together</em>
all_names = male_names.union(female_names)
#<em>Finds the names that appear in both lists, just returns those</em>
neutral_names = male_names.intersection(female_names)
#<em>Returns names that are NOT in both lists</em>
specific_names = male_names.symmetric_difference(female_names)
The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.
<h3>Ohm's law</h3>
Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as
V = IR
Where
V is the potential difference
I is the current
R is the resistance
<h3>Technician A</h3>
High resistance causes an increase in current flow
V = IR
Divide both side by I
R = V / I
Thus, technician A is wrong as high resistance suggest low current flow
<h3>Technician B</h3>
Higher than normal voltage drop could indicate high resistance
V = IR
Thus, technician B is correct as high voltage indicates high resistance
<h3>Conclusion </h3>
From the above illustration, we can see that technician B is correct
Learn more about Ohm's law:
brainly.com/question/796939
Answer:
B. Acid rain.
C. Photochemical smog.
Explanation:
Oxides of nitrogen contribute to the formation of photochemical smog and acid rain. Photochemical smog is a type of smog produced when ultraviolet light from the sun reacts with nitrogen oxides in the atmosphere while on the other hand, when nitrogen oxide react with the water vapor in the atmosphere forming nitric acid which falls on the earth surface with the help of precipitation.
