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2 years ago

15. Whether technology is good or bad depends on how it is used.

Engineering

2 answers:

Reptile [31]2 years ago

4 0

Answer:

true

Explanation:

miskamm [114]2 years ago

4 0

Answer:

True

Explanation:

The internet is a very useful tool, but remember there are billions of people out there who would use it for good purpose or bad ones.

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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

Joinn my zo om lets play some blookets<br> 98867 708157<br> 9dPQPW

pav-90 [236]

Answer:k

Explanation:

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2 years ago

Oxygen combines with nitrogen in the air to form NOx at about

rusak2 [61]

B) Oxygen combines with nitrogen in the air to form NOx at about 2500 degrees Fahrenheit.

4 0

3 years ago

Read 2 more answers

Describe, in a general form, the equation, in time domain, that tells the voltage across a inductor, L, as a function of time wh

love history [14]

Answer:

a) V(t) = Ldi(t)/dt

b) If current is constant, V = 0

Explanation:

a) The voltage, V(t), across an inductor is proportional to the rate of change of the current flowing across it with time.

If V represents the Voltage across the inductor

and i(t) represents the current across the inductor in time, t.

V(t) ∝ di(t)/dt

Introducing a proportionality constant,L, which is the inductance of the inductor

The general equation describing the voltage across the inductor of inductance, L, as a function of time when a current flows through it is shown below.

V(t) = Ldi(t)/dt ..................................................(1)

b) If the current flowing through the inductor is constant i.e. does not vary with time

di(t)/dt = 0 and hence the general equation (1) above becomes

V(t) = 0

4 0

3 years ago

PLEASE HELP QUICK!!

ivolga24 [154]

R01= 14.1 Ω

R02= 0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω