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Andru [333]
2 years ago
10

The alternator produces direct current, which is changed to alternating current by the diodes.

Engineering
1 answer:
exis [7]2 years ago
8 0

Answer:

False

Explanation:

<u>Alternat</u>or produces <u>alternat</u>ing current  ....diodes convert this to direct current

You might be interested in
In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab
Aliun [14]

Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

1.

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

2.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

Explanation:

7 0
3 years ago
The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c
marissa [1.9K]

Answer:

Code is given below:

Explanation:

.data  

str1: .space 20  

str2: .space 20  

msg1:.asciiz "Please enter string (max 20 characters): "  

msg2: .asciiz "\n Please enter string (max 20 chars): "  

msg3:.asciiz "\nSAME"  

msg4:.asciiz "\nNOT SAME"  

.text

.globl main

main:  

   li $v0,4        #loads msg1  

   la $a0,msg1  

   syscall

   li $v0,8

   la $a0,str1

   addi $a1,$zero,20

   syscall          #got string to manipulate

   li $v0,4        #loads msg2

   la $a0,msg2

   syscall

   li $v0,8

   la $a0,str2

   addi $a1,$zero,20

   syscall         #got string  

       la $a0,str1             #pass address of str1  

   la $a1,str2         #pass address of str2  

   jal methodComp      #call methodComp  

   beq $v0,$zero,ok    #check result  

   li $v0,4

   la $a0,msg4

   syscall

   j exit

ok:  

   li $v0,4  

   la $a0,msg3  

   syscall  

exit:  

   li $v0,10  

   syscall  

methodComp:  

   add $t0,$zero,$zero  

   add $t1,$zero,$a0  

   add $t2,$zero,$a1  

loop:  

   lb $t3($t1)         #load a byte from each string  

   lb $t4($t2)  

   beqz $t3,checkt2    #str1 end  

   beqz $t4,missmatch  

   slt $t5,$t3,$t4     #compare two bytes  

   bnez $t5,missmatch  

   addi $t1,$t1,1      #t1 points to the next byte of str1  

   addi $t2,$t2,1  

   j loop  

missmatch:    

   addi $v0,$zero,1  

   j endfunction  

checkt2:  

   bnez $t4,missmatch  

   add $v0,$zero,$zero  

endfunction:  

   jr $ra

3 0
3 years ago
What are three types of land reform ​
s344n2d4d5 [400]

Answer:

Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);

5 0
2 years ago
Read 2 more answers
HW6P2 (20 points) The recorded daily temperature (°F) in New York City and in Denver, Colorado during the month of January 2014
Maurinko [17]

Answer & Explanation:

function Temprature

NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];

DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];

%AVERAGE CALCULATION AND ROUND TO NEAREST INT

avgNYC=round(mean(NYC));

avgDEN=round(mean(DEN));

fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);

fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);

%part B

count=1;

NNYC=0;

NDEN=0;

while count<=length(NYC)

   if NYC(count)>avgNYC

       NNYC=NNYC+1;

   end

   if DEN(count)>avgDEN

        NDEN=NDEN+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);

fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);

%part C

count=1;

highDen=0;

while count<=length(NYC)

   if NYC(count)>DEN(count)

       highDen=highDen+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);

end

%output

check the attachment for additional Information

8 0
3 years ago
A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp
iragen [17]

Answer:

Elastic modulus of steel  = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm  = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × 10^{-4} m

solution

we know that Elastic modulus is express as

Elastic modulus = \frac{stress}{strain}    ................1

here stress is

Stress = \frac{Force}{area}       .................2

Area = (0.022)²

and  

Strain = \frac{extension}{length}       .............3

so here put value in equation 1 we get

Elastic modulus = \frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }  

Elastic modulus of steel = 202.27 × 10^9 Pa

Elastic modulus of steel  = 202.27 GPa

3 0
3 years ago
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