Solve the inequality below.Use the drop-down menus to describe the solution and its graph.

Ma puteti ajuta cu un argument de 2 pagini despre inlocuirea garniturii de etansare de pe pistonul etrierului de franare la un a

Nitella [24]

Answer:

can you translate

Explanation:

what Is that?

4 0

3 years ago

Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di

blsea [12.9K]

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

$\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )$

And $\tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )$

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

4 0

3 years ago

What is the function rule for the line? f(x)=−32x−2f(x)=−23x−2f(x)=32x−2f(x)=−32x+2A coordinate grid with x and y axis ranging f

murzikaleks [220]

Answer:

f(x)=23x−2

Explanation:

still trying to figure that out

7 0

3 years ago

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C

Qa= 5 MW

Lets heat supplied at 150°C = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

$COP=\dfrac{Qr}{W}$

$2.5=\dfrac{5}{W}$

W=2 MW

For Carnot heat pump

$COP=\dfrac{T_2}{T_2-T_1}$

$2.5=\dfrac{T_2}{T_2-(273+85)}$

2.5 T₂ - 895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0

3 years ago

LC3 Programming ProblemUse .BLKW to set up an array of 10 values, starting at memory location x4000, as in lab 4.Now programmati

irga5000 [103]

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

7 0

3 years ago