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3 years ago

PLEASE HURRY!!

Engineering

2 answers:

neonofarm [45]3 years ago

8 0

Atoms hope that helped

bezimeni [28]3 years ago

6 0

Answer:

Atoms hope this is helpful and good luck everyone Have a great day

Explanation:

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Joseph wants to practice architecture. Which compulsory assessment administered by NCARB does he need to complete?

Hunter-Best [27]

Answer:

Architect Registration Examination (ARE)

Explanation:

In this scenario, Joseph wants to practice architecture. A compulsory assessment administered by National Council of Architectural Registration Boards (NCARB) which he has to complete is an Architect Registration Examination.

An Architect Registration Examination (ARE) refers to the professional licensure examination to be taken by anyone who intends to practice architecture in the United States of America, Puerto Rico, Guam, Canada, US Virgin Islands.

The main purpose of the Architect Registration Examination is to assess an architect's knowledge, skills, and abilities on architectural best practices, procedures and services. Also, it focuses on areas relating to a building's safety, soundness and health impact on the habitants. Therefore, ARE is a prerequisite for practicing architecture across the United States of America jurisdiction.

Once an architect passes the examination, he or she would be given a license to practice architecture.

Hence, for Joseph to practice architecture he must take the Architect Registration Examination developed and administered by National Council of Architectural Registration Boards (NCARB).

8 0

3 years ago

A 100 MHz generator with Vg= 10/00 v and internal resistance 50 ohms air line that is 3.6m and terminated in a 25+j25 ohm load

Lilit [14]

The value of Vz at point z from the generator = 17.748 ∠ -107.62° V

Given data :

Internal resistance = 50 ohms

Vg = 10 ∠ 0° v

length of air line = 3.6 m

Terminating resistance = 25 + j25 Ω

First step : Determine the Total impedance 

Total Impedance ( z ) = Rin + Rline + Rl + jXl

= 50 + 50 + 25 + j25

= 125 + j25 ≈ 127.47 ∠ 11.3°

Next step : Determine the current in the circuit

current ( I ) = Vg / z

= ( 10∠0° v ) / ( 127.47 ∠ 11.3° )

= 0.0784 ∠ -11.3 amp

Final step : Determine the value of Vz at point z from the generator

Vz = ( Vg + I * Ri ) - ( RI * I + Rline * I )

= ( 10∠0° + 0.0784 ∠ -11.3 * 50 ) - ( 25 + j25 + 50 * 0.0784 ∠ -11.3 )

= -5.37 - j16.91 ≈ 17.748 ∠ -107.62° V

Hence we can conclude that the value of Vz at point z form the generator 17.748 ∠ -107.62° V

Learn more about voltage : brainly.com/question/11288970

Hello your question is incomplete below is the complete question

A 100 MHz generator with Vg =10< 0 degree (v) and internal resistance 50-ohm is connected to a lossless 50-ohm air line that is 3.6m long and terminated in a 25+j25 (ohm) load.

Find (a) V(z) at a location z from the generate.

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3 years ago

What is the relationship between the distance of the fall (height) and the formation of splines and satellite drops?

kondor19780726 [428]

Mammas mama,aka Kim’s,smamsmsmmsmsmsmsmsmsmamsmmsms,sms,sms,msms,sms,sms,sms,sis,s,s,s

7 0

3 years ago

Rivers account for _____% of the earth's fresh water. ?

Black_prince [1.1K]

Answer:

0.49%

Explanation:

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3 years ago

Read 2 more answers

A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identi

vladimir2022 [97]

Answer: the thermal conductivity of the sample is 22.4 W/m . °C

Explanation:

We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.

ASSUMPTIONS

1. Steady operating conditions exist since the temperature readings do not change with time.

2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.

3. The apparatus possess thermal symmetry

ANALYSIS

The electrical power consumed by resistance heater and converted to heat is:

Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W

Q = 1/2Wₐ = 1/2 ( 44 A )

Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so

A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²

Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be

Q = kA ( ΔT/L ) → k = QL / AΔT

k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )

k = 22.4 W/m . °C

3 0

3 years ago