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3 years ago

Your study space does not need to be quiet as long as you can ignore any noise coming from the space true or false?

Engineering

2 answers:

Andrews [41]3 years ago

5 0

Answer:

the answer is false

Explanation:

i got a 100 on edg 2021

Makovka662 [10]3 years ago

3 0

Answer:

False

Explanation:

When you're studying, you need to make sure that you can focus properly. This means that you shouldn't be hungry or too full and that you should be well-rested, in a quiet room with good lighting and no distractions. Noise is never good when you need to memorize something. Some people can partially ignore it as long as it isn't too loud, but it will begin to bother them eventually. That's why it's better to study in a quiet room.

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What is the inductive reactance of a 20 mH inductor at a frequency of 100Hz?

Inessa05 [86]

Answer:

The correct approach is "12.56 Ω".

Explanation:

The given values are:

Frequency,

f = 100 Hz

Inductor length,

L = 20 mH

Now,

The inductive reactance will be:

⇒ $X_L=2 \pi fL$

On putting the estimated values, we get

⇒ $=2 (3.14)\times 100\times 20\times 10^{-3}$

⇒ $=12.56 \ \Omega$

8 0

3 years ago

1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un

ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

Bearing life: $L_1 = L , L_2 = 2L$

Catalog rating: $C_1 = C , C_2 = ? ,$

From given equation bearing life equation,

$F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)$

we Dividing eqn (2) with (1)

$\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C$

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

$R_1 = 0.9 , R_2 = 0.99$

Now calculating life adjustment factor for both value of reliability from Weibull parametres

$a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}$

$= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968$

Similarly

$a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215$

Now calculating bearing life for each value

$L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L$

Now using given ball bearing life equation and dividing each other similar to previous problem

$\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C$

Catalog rating increased by factor of 0.61

6 0

3 years ago

4kj of energy are supplied to a machine used for lifting a mass.The force required is 800N.If the machine has an efficiency of 5

Bogdan [553]

I do not know shdjfjdjfk

7 0

3 years ago

Question 5 of 10

Angelina_Jolie [31]

The cubic inch space that is required inside a box for 4 #6 XHHN current carrying conductors will be D. 8 cubic inches.

<h3>What is a conductor?</h3>

It should be noted that a conductor simply means a substance or material that simply allows electricity to pass through it.

From the information given, the cubic inch space that is required inside a box for 4 #6 XHHN current carrying conductors is yo be computed.

This will be:

= (4 × 6)/3

= 24/3

= 8

In conclusion, the cubic inch space that is required inside a box for 4 #6 XHHN current carrying conductors will be 8 cubic inches.

Learn more about conductor on:

brainly.com/question/11845176

#SPJ1

6 0

2 years ago

Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater heater. Superheated vapor enters the

Oksanka [162]

Answer:

<u> solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment</u>

Download docx

3 0

3 years ago