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m_a_m_a [10]
3 years ago
7

If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen

d
Engineering
1 answer:
Mademuasel [1]3 years ago
6 0

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

Number of bits per seconds =  4 × 8000/second

Number of bits per seconds = 32000 bits/seconds

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This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and
vova2212 [387]

Answer:

A continuity test

Explanation:

A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the cord sets, receptacles that aren't part of a building or structure's permanent wiring, and cord-and-plug connected equipment required to be grounded. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.

Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment

8 0
3 years ago
For goods-producing firms, at which of the following levels of resource planning does scheduling for individual subassemblies an
VashaNatasha [74]

Answer:

Disaggregation

Explanation:

In a company it is a way to create operational plans that are focused, either by time or by section.

3 0
3 years ago
Which of the following is correct regarding the principal stresses and maximum in-plane shear stresses? a. Principal stresses ca
ziro4ka [17]

Answer:

option B.

Explanation:

The correct answer is option B.

Principal stress is the maximum normal stress a body can have. In principal stress, there is purely normal stress. On principal plane shear stress is zero.

In-plane shear stress are the shear stress which is acting on the plane.

The statement which is correct regarding principle plane and shear stress is that The shear stress over principal stress planes is always zero.

4 0
3 years ago
When the electrical connection to the alternator from the battery is not tight, it can cause?
Oksi-84 [34.3K]

Answer:

affects the flow of electricity

Explanation:

A loose battery terminal affects the flow of electricity. There is less power going to the electrical systems and the vehicle will not start or start sluggishly. Also, a loose battery terminal causes the car's electrical components like navigation, car lights, and audio among others to dim or fail completely.

3 0
2 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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