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3 years ago

The _____ is a voice for small business.

Engineering

1 answer:

Sliva [168]3 years ago

6 0

Answer:

The <u>National Federation of Independent Business (NFIB) </u>is a voice for small business

Explanation:

The National Federation of Independent Business is a not for profit business association with goal of supporting and upholding small business owners right to carry out their privately owned business, protecting them from excessive government intrusions into their businesses.

The association have their headquarters in Tennessee with offices in Washington DC and the capitals of the 50 state capitals making NFIB the most far-reaching and largest association for small business in the U.S.

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Which of the following best describes a central idea of the text?

GREYUIT [131]

Answer:

i think is B

Explanation:

6 0

3 years ago

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

GrogVix [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$\frac{1}{C_{eq}} = \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47}$

$\frac{1}{C_{eq}} = \frac{8620}{517}$

$C_{eq} = \frac{517}{8620}$

$C_{eq} = 0.0599$

Rounding off yields

$C_{eq} = 0.060 \: \mu F$

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

5 0

3 years ago

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

An approach to a signalized intersection has a saturation flow rate of 1800 veh/h. At the beginning of an effective red, there a

Tcecarenko [31]

Answer: The total vehicle delay is

39sec/veh

Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.

Effective red time (r) = 25sec

Arrival rate (A) = 900veh/h = 0.25veh/sec

Departure rate (D) = 1800veh/h = 0.5veh/sec

STEP1: FIND THE TRAFFIC INTENSITY (p)

p = A ÷ D

p = 0.25 ÷ 0.5 = 0.5

STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE

The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.

Dt = (A × r^2) ÷ 2(1 - p)

Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)

Dt = 156.25 ÷ 4 = 39.0625

Therefore the total vehicle delay after one cycle is;

Dt = 39

4 0

3 years ago

La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación

gregori [183]

Answer:

I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

I₀ = $\frac{I}{sin \ \theta }$

I₀ = 1600 /sin 53

I₀ = 2003.42 lx

for when the angle is θ = 37º

I = 2003.42 sin 37

I = 1205.69 Lx

6 0

3 years ago