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2 years ago

Valorant Or Csgo? which is best in your opinion?

Engineering

2 answers:

SashulF [63]2 years ago

7 0

Answer:

valorant valorant valorant is the best

masya89 [10]2 years ago

4 0

Valorant is obviously the better one here.

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Input Energy ---> Output Energy

uranmaximum [27]

Answer:

motion ------> electrical. winds push the turbines which generate a magnetic fields which in turn, generates electricity

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2 years ago

In which situation is a are food service workers not required to wash their hands?

Margarita [4]

Answer:

when wearing gloves?

Explanation:

or when off duty

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2 years ago

Read 2 more answers

An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is

Darina [25.2K]

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

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

What new information about hte role of the y subunit of ATP Synthase did this researcdh project elucidate? how does the subunit

nignag [31]

Answer:

Explanation:

it is observed given the information by the y subunit of ATP synthase that when one variable is changed, F1 motor becomes reversible and this is achieved by having the subunit as the reversible motor.

4 0

3 years ago