1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Scilla [17]
3 years ago
5

Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s

treet repair area of 2000 ft by 48 ft by 6 in. The aggregate in the stockpile contains 3.1 percent moisture. If the required compaction is 95% of the target, how many tons of aggregate will be needed?
Engineering
1 answer:
djyliett [7]3 years ago
3 0

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

You might be interested in
Thermal energy measured by?
Harlamova29_29 [7]

Answer:

Thermal energy is measured using a thermometer denominated in Fahrenheit, Celsius and Kelvin

3 0
2 years ago
Lynx eat snowshoe hares, and snowshoes hears eat plants. Which term can be applied to the lynx in this food chain example? Prima
erastova [34]

Answer:

primary consumer because YES

3 0
2 years ago
A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur
Lina20 [59]

Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and  increasing the capacitance proportionally to the dielectric constant of the Teflon.  

8 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
A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,
disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

4 0
3 years ago
Other questions:
  • Use the drop-down menus to choose the correct term or words to complete the statements.
    10·1 answer
  • What is something that a robot or computer program might do that requires a decision, or conditional statement?
    9·1 answer
  • What is the main role of matrix in composites! a)-to transfer stress to the other phases b)- to protect phases from environment
    7·1 answer
  • Which of the following drivers has the right-of-way?
    9·1 answer
  • When testing a compressor with an ohm meter, a technician read 2 ohms between the start terminal and the case of the compressor.
    5·1 answer
  • How can goal setting help with academic performance?
    13·1 answer
  • What ic engine for mechanic
    15·1 answer
  • How to change a fuel fiter
    12·1 answer
  • In the car industry, clay models are used to visualize and test new car designs.
    5·2 answers
  • Ignition for heavy fuel oil?
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!