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3 years ago

ANSWER FAST PLEASE!!! WILL MARK BRAINLIEST!!!!!!

Engineering

2 answers:

Phoenix [80]3 years ago

5 0

Answer:

Phuong works on a research project and creates a report for her boss.

vodka [1.7K]3 years ago

4 0

Answer:

B. Bart types information into a computer database.

C. Phuong works on a research project and creates a report for her boss.

D. Anton answers phone calls and greets guests who visit a company.

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The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

Milk has a density of as much as 64.6 lb/ft3. What is the gage pressure at the bottom of the straw 6.1 inches deep in the milk?

gregori [183]

Answer:

Explanation:

1 inch is 0.0833333feet

6.1 inches is 0.5083 feet

Density = mass/volume

64.6 = mass/0.50833

mass = 64.6 x 0.5083 =32.83618lb

3 0

3 years ago

Which of the following is NOT true about hydraulic systems?

Dmitry [639]

Answer: The answer is D

D.In hydraulic systems, the operating temperatures must be kept between 170�F and 180°F

Explanation:

The operating temperature for hydraulic systems is 140°F and below. Anything above this temperature is too high and will reduce the useful life of hydraulic fluid.

Most often problems associated with hydraulic systems are caused by fluid contaminated with particulate matter.

7 0

3 years ago

A 4-kW electric heater runs for 2 hours to raise the room temperature to the desired level. Determine the amount of electric ene

Anna35 [415]

Answer:

Q' = 8 KW.h

Q'=28800 KJ

Explanation:

Given that

Heat Q= 4 KW

time ,t = 2 hours

The amount of energy used in KWh given as

Q ' = Q x t

Q' = 4 x 2 KW.h

Q' = 8 KW.h

We know that

1 h = 60 min = 60 x 60 s = 3600 s

We know that W = 1 J/s

The amount of energy used in KJ given as

Q' = 8 x 3600 = 28800 KJ

Therefore

Q' = 8 KW.h

Q'=28800 KJ

6 0

3 years ago

1. Assume that EAX contains ff ff ff 51 and the doubleword referenced by value contains ff ff ff f1.

IRINA_888 [86]

Answer:

1. True

2. False

Explanation:

given data

EAX contains = ff ff ff 51

doubleword referenced = ff ff ff f1

conditional jump add = eax

solution

1st statement is true

but 2nd statement is false

as here

js or jne instruction is the conditional jump that is follow a test
It jump to the specified location when previous instructions are set the SF (Sign Flag) .

6 0

3 years ago