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2 years ago

You are coming to this intersection, and are planning on turningright. There is a vehicle close behind you. You should?

Engineering

1 answer:

Advocard [28]2 years ago

3 0

Answer:

Put on your right turn signal

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if a current of 5 amps flows through a resistance of 40 ohms, what is the voltage across that resistor

olasank [31]

Answer:

200V

Explanation:

I = 5A

R = 40Ω

V = IR = (5)(40) = 200V

7 0

2 years ago

List in order first three steps to square a board

LenaWriter [7]

Answer:

STEP1 Cut to Rough Length

STEP2 Cut to Rough Width

STEP 3 Face-Jointing

HOPE THAT HELPSSS!!!

5 0

2 years ago

Suppose you were doing an A* search and there were a number of different ways that you could compute an admissible heuristic val

Elodia [21]

Answer:

Explanation:

First of all,we need to know what kind of research and how important it is before determining which computation method to use .

Then compare end result of research and cost of expensive methods .

Then we should opt for the best expensive method to be used for the given application.

5 0

2 years ago

One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isotherm

maria [59]

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a) Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

6 0

2 years ago

Explain the following statement: In a windowless proportional counter, the output pulse height from an alpha particle source wil

-BARSIC- [3]

Answer / Explanation:

Regarding α and β Particles in Windowless Counter, the range of α particles is lower than β particles. Alpha particles typically have range less than the dimensions of the gas chamber so that proportional counters are able to easily record. Hence, with almost 100% efficiency, each particle which enters the so called active volume.

Then, since the pulse height spectra depends on the number of ion pairs which have formed, an aplha particle with higher energy creates more ion pairs in the chamber. However, beta particle range usually exceeds the dimensions of the chamber and therefore most of the betas hit the walls where they deposit energy. Then, fewer ion pairs are formed because very few β’s give their energy to the bulk gas.

5 0

3 years ago