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borishaifa [10]
3 years ago
9

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:
Mice21 [21]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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meriva
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Sump heater

Explanation
3 0
3 years ago
Chapter 15 – Fasteners: Determine the tensile load capacity of a 5/16 – 18 UNC thread and a 5/16 – 24 UNF thread made of the sam
Roman55 [17]

Answer:

The 5/16 – 24 UNF is stronger because it has more tensile load capacity.

Tensile load capacity for M8 -1.25 = 5670 lb

Tensile load capacity for M8 -1 = 6067 lb

Explanation:

For 5/16 - 18 UNC thread:

D = 0.3125

n = 18

Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18) ^2

= 5243 lb.

Similarly for 5/16 - 24 UNF , only the n value changes to 24

we get the tensile load capacity = 5806.6 lb

Hence the 5/16 – 24 UNF is stronger because it has more tensile load capacity.

For metric Bolts:

We have to consider all values in SI units

Strength = 689 MPa

We get for M8 -1.25:

Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb

For M8 -1:

Tensile load capacity as = 689 X 39.167 = 26986 N = 6067lb

7 0
2 years ago
Read 2 more answers
Hỗ trợ mình với được không các bạn
Leya [2.2K]

Answer:

Explanation:

Be bop

6 0
3 years ago
Fixed rate mortgage offer:
jarptica [38.1K]

Answer: The answer is :

      C) $887

Explanation:

5 0
3 years ago
The diffusion coefficients for species A in metal B are given at two temperatures:
Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

D = D_0e^{\frac{-Q_d}{RT} }

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0
3 years ago
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