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pickupchik [31]
2 years ago
7

Before you disconnect the service battery from the discharged battery, it is good practice to place a load across the

Engineering
2 answers:
Lilit [14]2 years ago
6 0

Answer:

it is true i just did this test

Explanation:

alexgriva [62]2 years ago
3 0
The answer should be true
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A shunt regulator utilizing a zener diode with an incremental resistance of 8 ohm is fed through an 82-Ohm resistor. If the raw
spayn [35]

Answer:

\triangle V_0=0.08V

Explanation:

From the question we are told that:

Incremental resistance  R=8ohms

Resistor Feed R_f=82ohms

Supply Change \triangle V=1

Generally the equation for  voltage rate of change is mathematically given by

 \frac{dV_0}{dV}=\frca{R}{R_1r_3}

Therefore

 \triangle V_0=\triangle V*\frac{R}{R_fR}

 \triangle V_0=1*\frac{8}{8*82}

 \triangle V_0=0.08V

7 0
3 years ago
How can I draw this image in 2D form
Ket [755]

Answer:

no it is not 2D

Explanation:

it is 3D

ok so follow these steps

- make hole

-make square

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ok now your figure is ready

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Q1: You have to select an idea developing an application like web/mobile or industrial, it should be based on innovative idea, n
elena55 [62]

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Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.

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if I am right then make me brainliest

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3 years ago
Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic
Tom [10]

Answer:

Explanation:

The pictures below shows the whole explanation for the problem

4 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
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