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alina1380 [7]
3 years ago
6

Which of the following describes a benefit of monitoring sensor values in a VR Project?

Engineering
2 answers:
inysia [295]3 years ago
6 0
The answer will be a
kvv77 [185]3 years ago
4 0
The answer will be a
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The dry unit weight of a soil sample is 14.8 kN/m3.
Jlenok [28]

Answer:

See attachment for completed question

Explanation:

Given that; Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(c) Degree of saturation

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

4 0
4 years ago
You are an engineer working in a auto crash test lab. Some members of your team have raised objections against the use of cadave
Nikolay [14]

Answer: Application.

Explanation:

The question on wether to contine the use of cadavers in the lab for test is being centered around its application. Cadaver which is same as a corpse or dead body is used in crash site during automobil test in lab, some of this cadavers are been disrespected with their applications in the automobile industries because many didn’t consent to be used in those experiments or test.

5 0
3 years ago
There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.
sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})

C_{1}=1.2606\times10^{-5} $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})

C_{2}=3.334\times10^{-5} $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})

C_{3}=1.66\times10^{-5} $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

 

5 0
3 years ago
A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle
Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

X_{s} =1.23(\frac{V_{line}^{2}  }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms

the network voltage phase is

V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV

the power transmitted is equal to:

|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV

the line induced voltage is

|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV

7 0
3 years ago
A __________ is an added note showing additional or more specific information.
xxMikexx [17]

Answer:

awnsers should be added to know to show additional

8 0
3 years ago
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