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3 years ago

Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.

Engineering

1 answer:

likoan [24]3 years ago

4 0

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

print("Largest: " + string(number2))

else:

print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))

# END #

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A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second

Andreyy89

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.

4 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

Explicar el funcionamiento de un multímetro analógico.

Whitepunk [10]

Answer:

Un multímetro analógico funciona como un medidor de bobina móvil de imán permanente (PMMC) para tomar mediciones eléctricas

Explanation:

El multímetro analógico es un medidor o galvanómetro D'Arsonval que funciona según el principio de los medidores de bobina móvil de imán permanente (PMMC)

Un multímetro analógico está formado por un puntero de aguja unido a una bobina móvil colocada entre el polo norte y sur de un imán permanente dispuesto de tal manera que, cuando una corriente eléctrica fluye a través de la bobina, genera una fuerza de campo magnético que interactúa con el imán fuerza de campo de los imanes permanentes que hace que la bobina se mueva junto con el puntero de la aguja sobre un dial graduado

Para controlar el movimiento del puntero de la aguja, de modo que el par requerido para producir una cantidad de movimiento por corriente detectada por el multímetro, se colocan dos resortes a través de la bobina para proporcionar resistencia al movimiento en ambas direcciones y para permitir la calibración del multímetro analógico.

4 0

3 years ago

At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide

Xelga [282]

Answer:

a) aA = - 13.33 mm/s²

aB = - 20 mm/s²

b) aD = - 13.33 mm/s²

c) vB = 70 mm/s

d) xB = 440 mm

Explanation:

Given

The initial speed of B is: v₀B = 150 mm/s

Distance moved by A is: xA = 240 mm

Velocity of A is: vA = 60 mm/s

Assuming:

Displacement of blocks are denoted by:

A = xA

B = xB

C = xC

D = xD

From the pic shown, the total length of the cable is:

xB + (xB - xA) + 2*(d - xA) = L

⇒ 2*xB - 3*xA = L - 2*d

where L - 2*d is constant. Differentiating the above equation with respect to time:

d(2*xB)/dt - d(3*xA)/dt = 0

⇒ 2*vB - 3*vA = 0 (i)

Substituting in equation (i)

2*(150 mm/s) - 3*vA = 0

⇒ v₀A = 100 mm/s (initial speed of A)

Then, we use the equation

vA² = v₀A² + 2*aA*xA

Substituting the values in above equation:

(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)

⇒ aA = - 13.33 mm/s²

If 2*vB - 3*vA = 0

Differentiating the above equation with respect to time:

d(2*vB)/dt - d(3*vA)/dt = 0

⇒ 2*aB - 3*aA = 0 (ii)

Substituting in equation (ii)

2*aB - 3*(- 13.33 mm/s²) = 0

⇒ aB = - 20 mm/s²

b) From the pic shown,

xD - xA = constant

If we apply

d(xD)/dt - d(xA)/dt = 0

⇒ vD - vA = 0

then

d(vD)/dt - d(vA)/dt = 0

⇒ aD - aA = 0

⇒ aD = aA = - 13.33 mm/s²

c) We use the formula

vB = v₀B + aB*t

Substituting the values in above equation:

vB = 150 mm/s + (- 20 mm/s²)*(4 s)

⇒ vB = 70 mm/s

d) We apply the equation

xB = v₀B*t + 0.5*aB*t²

Substituting the values in above equation:

xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²

⇒ xB = 440 mm

4 0

3 years ago

The mechanical advantage of a screw is always ____________________ than/to 1. Question 5 options: less, greater, equal, none of

torisob [31]

Answer:well u can use to make a shelter but that's all I can think of ??

Explanation:

3 0

3 years ago