Answer:
A tunnel built in soft ground—such as clay, silt, sand, gravel or mud—requires specialized techniques compared to hard rock, to compensate for the shifting nature of the soil.
It's from web...
Soft ground tunneling describes the additional measures needed when Microtunneling through soil conditions that are vulnerable to collapse. ... This process ensures tunneling can happen effectively in soft grounds.
It's from me...
An alternating copolymer is known to have a number-average molecular weight of 100,000 g/mol and a degree of polymerization of 2210. If one of the repeat units is ethylene, which of styrene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit?
a. vinyl chloride
b. propylene
c. tetrafluoroethylene
d. styrene
Answer:
d. Styrene
Explanation:
An alternating copolymer repeat unit types if the other is Styrene. The equation to calculate m is :
m = Mn / Dp
m = 100,000 / 2210 = 45.25g/ mol
alternating copolymer is chain fraction than of each repeat unit type then chain fraction is styrene and unknown repeat units.
A viscous liquid is sheared between two parallel disks; the upper disk rotates and the lower one is fixed. The velocity field between the disks is given by V=e^θrωz/h (The origin of coordinates is located at the center of the lower disk; the upper disk is located at z = h.) What are the dimensions of this velocity field? Does this velocity field satisfy appropriate physical boundary conditions? What are they?
Answer:
For lower disk : V = e^θrω(0)/h = 0
At the upper disk: V = e^θrω(h)/h = e^θrω
Hence The physical boundary conditions are satisfied
Explanation:
Velocity field ( V ) = e^θrωz/h
Upper disk located at z = h
Determine the dimensions of the velocity field
velocity field is two-dimensional ; V = V( r , z )
applying the no-slip condition
condition : The no-slip condition must be satisfied
For lower disk Vo = 0 when disk is at rest z = 0
∴ V = e^θrω(0)/h = 0
At the upper disk V = e^θrω given that a upper disk it rotates at z = h
∴ V = e^θrω(h)/h = e^θrω
Hence we can conclude that the velocity field satisfies the appropriate physical boundary conditions.
The view factor for radiation emitted by surface 1 to surface 2 was calculated to be 0.4. The working area of surface 1 is 0.01 m2, the working area of surface 2 is 0.04 m2. What is the view factor for radiation emitted by surface 2 to surface 1?
Answer:
The view factor for radiation emitted by surface 2 to surface 1 is 0.1
Explanation:
Given
[tex]F_{12} = 0.4[/tex]
[tex]A_1 = 0.01m^2[/tex]
[tex]A_2 = 0.04m^2[/tex]
Required
Determine [tex]F_{21}[/tex]
To do this, we make use of the following equivalent ratio
[tex]A_1 * F_{12} = A_2 * F_{21}[/tex]
Make [tex]F_{21[/tex] the subject
[tex]F_{21} = \frac{A_1 * F_{12}}{ A_2}[/tex]
Substitute values into the equation
[tex]F_{21} = \frac{0.01m^2 * 0.4}{0.04m^2}[/tex]
[tex]F_{21} = \frac{0.01 * 0.4}{0.04}[/tex]
[tex]F_{21} = \frac{0.004}{0.04}[/tex]
[tex]F_{21} = 0.1[/tex]
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles of the gas are present in the tank? What is the molecular weight of the gas? Assuming that the gas to be a pure element can you identify it?
Answer:
The molecular weight will be "28.12 g/mol".
Explanation:
The given values are:
Pressure,
P = 10 atm
= [tex]10\times 101325 \ Pa[/tex]
= [tex]1013250 \ Pa[/tex]
Temperature,
T = 298 K
Mass,
m = 11.5 Kg
Volume,
V = 1000 r
= [tex]1 \ m^3[/tex]
R = 8.3145 J/mol K
Now,
By using the ideal gas law, we get
⇒ [tex]PV=nRT[/tex]
o,
⇒ [tex]n=\frac{PV}{RT}[/tex]
By substituting the values, we get
[tex]=\frac{1013250\times 1}{8.3145\times 298}[/tex]
[tex]=408.94 \ moles[/tex]
As we know,
⇒ [tex]Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}[/tex]
or,
⇒ [tex]MW=\frac{m}{n}[/tex]
[tex]=\frac{11.5}{408.94}[/tex]
[tex]=0.02812 \ Kg/mol[/tex]
[tex]=28.12 \ g/mol[/tex]
An L2 steel strap having a thickness of 0.125 in. and a width of 2 in. is bent into a circular arc of radius 600 in. Determine the maximum bending stress in the strap.
Answer:
the maximum bending stress in the strap is 3.02 ksi
Explanation:
Given the data in the question;
steel strap thickness = 0.125 in
width = 2 in
circular arc radius = 600 in
we know that, standard value of modulus of elasticity of L2 steel is; E = 29 × 10³ ksi;
Now, using simple theory of bending
1/p = M/EI
solve for M
Mp = EI
M = EI / p ----- let this be equation 1
The maximum bending stress in the strap is;
σ = Mc / I -------let this be equation 2
substitute equation 1 into 2
σ = ( EI / p)c / I
σ = ( c/p )E
so we substitute in our values
σ = ( (0.125/2) / 600 )29 × 10³
σ = 0.00010416666 × 29 × 10³
σ = 3.02 ksi
Therefore, the maximum bending stress in the strap is 3.02 ksi
A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses heat to atmosphere at 300 K. Assuming the block to be at steady state, determine (a) the net rate of heat transfer in kW; (b) the rate of entropy generation in the system's universe
Answer:
a) Zero
b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K
Explanation:
a) In steady state
Net rate of Heat transfer = net rate of heat gain - net rate of heat lost
Hence, the rate of heat transfer = 0
b) In steady state, entropy generated
ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]
Substituting the given values, we get –
ds/dt = -[5/1500 + 3/1000 – (5+3)/300]
ds/dt = - [0.0033 + 0.003 -0.2666]
ds/dt = 0.2603 KW/K
A uniform plane electromagnetic wave propagates in a lossless dielectric medium of infinite extent. The electric field in the wave has the instantaneous expression
E(r,t) = (ix √3 - iz) 2 sin(2π.10^8t + 2πx/3 + 2nz/√3 + 30 ), V/m.
Find:
a. iE, the unit vector in the direction of the wave electric field
b. the amplitude Eo of the wave
c. the wavelength of the wave
d. ik, the unit vector in the direction of propagation
Answer:
Explanation:
From the information given:
The instantaneous expression of the electric field in the wave is:
[tex]E(r,t)= (i_x \sqrt{3} -i_z) 2 \ sin (2 \pi*10^8t + 2 \pi x/3+2 \pi z /\sqrt{3} + 30 ^0) , \ V/m[/tex]
To determine the unit vector in line with the wave electric field, we take the first term in E(r,t) for [tex]I_E^\to[/tex] as:
[tex]I_E^\to = i_x \sqrt{3}-i_z \\ \\ I_E^\to = \dfrac{i_x \sqrt{3}-i_z}{\sqrt{3 +1}} \\ \\ \mathbf{ I_E = \dfrac{i_x\sqrt{3} -i_z}{2}}[/tex]
The amplitude is denoted by the numerical value after the first term, which is:
[tex]\mathbf{E_o = 2}[/tex]
The wavelength can be determined by using the expression:
[tex]\beta =\dfrac{2 \pi}{\lambda }[/tex]
from the given instantaneous expression:
[tex]\beta = \dfrac{2 \pi}{3}x+\dfrac{2 \pi}{\sqrt{3}}z[/tex]
[tex]\beta = \sqrt{\dfrac{2 \pi}{(3)^2}+\dfrac{(2 \pi}{(\sqrt{3})^2}}[/tex]
[tex]\beta = \sqrt{\dfrac{2 \pi}{9}+\dfrac{2 \pi}{{3}}}[/tex]
Factorizing 2π
[tex]\beta =2 \pi \sqrt{\dfrac{1}{9}+\dfrac{1}{{3}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{9+3}{9*3}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{12}{27}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{4*3}{9*3}}}[/tex]
[tex]\beta =2 \pi \sqrt{\dfrac{4}{9}}}[/tex]
[tex]\beta =2 \pi\times {\dfrac{2}{3}}}[/tex]
recall from the expression using in calculating wavelength:
[tex]\beta =\dfrac{2 \pi}{\lambda }[/tex]
∴
equating both together, we have:
[tex]\dfrac{2 \pi}{\lambda }= 2 \pi\times {\dfrac{2}{3}}}[/tex]
[tex]\lambda = \dfrac{3}{2}[/tex]
λ = 1.5 m
In line with the wave direction; unit vector [tex]i_k[/tex] can be computed as follows:
[tex]i_k = - [ \beta_1x +\beta_2z]/\beta[/tex]
where;
[tex]\beta_1 = \dfrac{2 \pi }{3} \ ; \ \beta_2 = \dfrac{2 \pi }{\sqrt{3}} \ ; \ \beta = \dfrac{2 \pi \times 2}{3} ;[/tex]
∴
[tex]i_k = - \Big[\dfrac{2 \pi}{3}x + \dfrac{2 \pi}{\sqrt{3}} z\Big]\times \dfrac{1}{\dfrac{2 \pi *2}{3}}[/tex]
[tex]i_k = - \Big[\dfrac{x}{2} + \sqrt\dfrac{{3}}{4}} z\Big][/tex]
[tex]i_k = - \Big[\dfrac{1}{2}x + \sqrt{\dfrac{3}{4} }z\Big][/tex]
[tex]\mathbf{i_k = - \Big[0.5x +0.86 z\Big]}[/tex]
convert 25 inches / min to mm/hour
Answer:
25 mins into hours = 0.416667 hours
25 inches as mm = 635
Explanation:
nơi nào có điện tích thì xung quanh điện tích đó có :
Explanation:
sory sorry sorry sorrysorrysorry
A school teacher must schedule seven sessions, which are abbreviated M, N, O, P, S, T, and U, during a day. Seven different consecutive time periods are available for the sessions, and are numbered one through seven in the order that they occur. Only one session can be schedules for each period. The assignment of the sessions to the periods is subject to the following restrictions:
M and O must occupy consecutive periods. M must be scheduled for an earlier period than U.
O must be scheduled for a later period than S.
If S does not occupy the fourth period, then P must occupy the fourth period.
U and T cannot occupy consecutively numbered periods.
Which of the following could be true?
a. M is assigned to the first period.
b. O is assigned to the fifth period.
c. S is assigned to the seventh period.
d. T is assigned to the sixth period.
Given:
There are seven sessions to be scheduled in seven different consecutive time periods. Only one session can be scheduled for each period. The sessions are abbreviated as M, N, O, P, S, T, U. The restrictions are:
(i) M & O must occupy consecutive periods.
(ii) M must be scheduled for an earlier period than U.
(iii) O must be scheduled for a later period than S.
(iv) If S does not occupy the fourth period, then P must occupy the fourth period.
(v) U & T cannot occupy consecutively numbered periods.
Solution:
We will construct the sequence of sessions based on the given restrictions.
Since M & O must occupy consecutive periods, we can have the sequence as {..., M, O, ...} or {..., O, M, ...}
Since M must be scheduled for an earlier period than U, we can have the sequence as {..., M, O, ..., U, ...} or {..., O, M, ..., U, ...}
Since O must be scheduled for a later period than S, we can have the sequence as {..., S, ..., M, O, ..., U, ...} or {..., S, ..., O, M, ..., U, ...}
We can see that, according to the given restrictions, M cannot be assigned the first period as S has to be assigned before M. Thus option (a) is incorrect.
We can see that, according to the given restrictions, S cannot be assigned to the seventh period as seventh period is the last period and M, O & U has to be assigned after S. Thus option (c) is incorrect.
Now, T can be assigned in the following ways:
(I) After U: In this case, there are at least 4 sessions before T, the last of which is U. Moreover, according to the given restrictions, U & T cannot occupy consecutive periods. Also, since we are assuming that S is the first element, the fourth element has to be P, so that U is assigned to 5th period or after. Thus T has to be assigned to 7th, if we skip the period after U. That is, T cannot be assigned to the 6th period in this case.
(II) Between O, M & U: Even if U is assigned to the last (7th) period, since U & T cannot occupy consecutive periods, T cannot be assigned the 6th period in this case.
(III) Between S & O, M: This would imply that there are at least 3 sessions after T. This would automatically imply that T cannot be assigned to the 6th period in this case.
(IV) Before S: This implies that there are at least 4 sessions after T. Thus, T cannot be assigned to the 6th period in this case either.
Thus, T cannot be assigned to the sixth period in any case. That is, option (d) is incorrect.
Now, following all the given restrictions, one of the arrangements can be,
{1-N, 2-P, 3-T, 4-S, 5-O, 6-M, 7-U}
We can see that S is occupying the 4th period & U and T are not occupying consecutive periods. Thus, all the restrictions are followed. We can see that it is possible for O to be assigned to fifth period by following all the restrictions. Thus option (b) is the correct choice.
Final answer:
Option (b) is the correct choice. That is, based on the given restrictions, O can be assigned to the fifth period.
The option that is true as regards the 7 sessions for the consecutive time periods under the given conditions is;
B; O is assigned to the fifth period.
We are given the seven sessions during the day as;
M, N, O, P, S, T and U.
There are seven consecutive time periods for the sessions with the following conditions;
Only one session can be schedules for each period.M and O must occupy consecutive periods.M must be scheduled for an earlier period than U.O must be scheduled for a later period than S. If S does not occupy the fourth period, then P must occupy the fourth period. U and T cannot occupy consecutively numbered periods.Combining the 2nd and third conditions above, we have the order;M, O, U or O, M, U.
Considering the fourth condition given with the order above, we have; S, M, O, U or S, O, M, U.Considering the fifth condition given with the orders above, we have the orders;- S, M, O, P, U
- T/U
, S
, N
, P
, T/U
, M/O
, M/O
- S
, T/U
, N
, P
, T/U
, M/O
, M/O
- S, O, M, P, U
Now, from the sixth condition we can say that the order for N, T and U is;U, N, T or T, N, U
Finally, looking at the options and considering the orders from the conditions applied, the only correct answer is that O can be assigned to the fifth period since M/O are in the fifth and sixth period from our arranged orders.Read more about Logical reasoning at; https://brainly.com/question/14458200
list 3 appliances each of the following
soil
waste
Answer:
soil>>Slop sink
>Urinal
>water closet
The sample calculation for iron oxide in the IDEAS section of this experiment used known atomic weights to calculate an empirical formula. However, early chemists did not have any references in which they could look up atomic weights. Instead, they guessed at the formulas of compounds and measured the percent compositions of elements in compounds in order to calculate atomic weights. Calculate an atomic weight for iron using the hypothetical formula Fe101 and the composition data given in the example in the IDEAS section. You may assume the atomic weight of oxygen is known from other sources to be 16 amu.
Answer:
37.33 grams
Explanation:
The missing information embedded in the idea section is attached in the image below:
The aim of this question is to determine the atomic wt. of Iron (Fe) from the hypothetic formula:
Fe₁O₁
Here, we know that the mole ratio can be written as:
[tex]\dfrac{O}{Fe}=\dfrac{1}{1}[/tex]
Suppose we assume that the atomic wt. of Fe = β(unknown)???
Then the grams of O and Fe that is contained in Fe₁O₁ can be expressed as:
For O:
1 × 16 grams of Oxygen = 16 grams of O
For Fe:
1 × β grams of Fe = β grams of Fe
Now, let's take a look at the idea experiment, the mole solution can be computed as:
[tex]\dfrac{O}{Fe} = \dfrac{3}{2} \\ \\ \text{It implies that} \implies \dfrac{(3\times 16) \text{grams of O}}{(2 \times 56 ) \ \text{grams of Fe}}[/tex]
Equating both expressions above, we have:
[tex]\implies \dfrac{16}{ \beta} = \dfrac{3\times 16}{2\times 56}[/tex]
[tex]{ \beta} = \dfrac{(2\times 56)\times 16}{ 3\times 16}[/tex]
[tex]\mathbf{{ \beta} = 37.33 \ grams}[/tex]
Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00
per hour for every hour worked above 40 hours. Assume that employees do not work for fractional
part of an hour.
Answer:
Here is the code.
Explanation:
#include<stdio.h>
int main()
{
int i, time_worked, over_time, overtime_pay = 0;
for (i = 1; i <= 10; i++)
{
printf("\nEnter the time employee worked in hr ");
scanf("%d", &time_worked);
if (time_worked>40)
{
over_time = time_worked - 40;
overtime_pay = overtime_pay + (12 * over_time);
}
}
printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);
return 0;
}
Output :
Enter the time employee worked in hr 42
Enter the time employee worked in hr 45
Enter the time employee worked in hr 42
Enter the time employee worked in hr 41
Enter the time employee worked in hr 50
Enter the time employee worked in hr 51
Enter the time employee worked in hr 52
Enter the time employee worked in hr 53
Enter the time employee worked in hr 54
Enter the time employee worked in hr 55
Total Overtime Pay Of 10 Employees Is 1020.
Do you know anything about Android graphics?
Android provides a huge set of 2D-drawing APIs that allow you to create graphics.
Android has got visually appealing graphics and mind blowing animations.
The Android framework provides a rich set of powerful APIS for applying animation to UI elements and graphics as well as drawing custom 2D and 3D graphics.
Three Animation Systems Used In Android Applications:-1. Property Animation
2. View Animation
3. Drawable Animation
It is essential to wait until the end of the project to check if the sponsor/customer requirements and expectations have been met regarding the quality of the project deliverables.
T/F
Answer:
False.
Explanation:
Project management can be defined as the process of designing, planning, developing, leading and execution of a project plan or activities using a set of skills, tools, knowledge, techniques and experience to achieve the set goals and objectives of creating a unique product or service.
Generally, projects are considered to be temporary because they usually have a start-time and an end-time to complete, execute or implement the project plan.
The fundamentals of Project Management includes;
1. Project initiation.
2. Project planning.
3. Project execution.
4. Monitoring and controlling of the project.
5. Adapting and closure of project.
Basically, the specifications of a project or manufacturing process outlines the minimum requirements and quality that are acceptable. Thus, it must be adhered to strictly in order to achieve a successful and desired outcome.
As a rule, it is essential to check at every stage of a project if the sponsor or customer requirements and expectations have been met regarding the quality of the project deliverables.
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 50%. To what height will it lift the mass?
What must you do when you reach a steady yellow traffic light?
Answer:
When you come up on a steady yellow traffic light you should always yield to cross traffic if you can yield safely. The flashing yellow light is there to inform drivers to be careful and to slow down.
Explanation:
hope it helped!
A confined aquifer with a transmissivity of 300 m2/day and a storativity of 0.0005 and a well radius of 0.3 m. Find the drawdown in the well at 100 days if the following pumping schedule is followed after a long period of time of no pumping.
Period
1 2 3 4
Time (days) 0-20 20-50 50-90 90-100
Q (m3/day) 500 300 800 0
Answer:
8.4627 m
Explanation:
Transmissivity( T ) = 300 m^2/day
Storativity( S ) = 0.0005
well radius ( r ) = 0.3m
Determine the drawdown in well at 100 days
Drawdown at 100 days = ∑ Drawdown at various period
We will use the equation : S = Q / U*π*T [ -0.5772 - In U ] ----- ( 1 )
where : Q = discharge , T = transmissivity
S = drawdown ,
U = r^2*s / 4*T*t --- ( 2 )
r = well radius , S = Storativity, t = time period
i) During 0-20
U1 = r^2*s / u*π*t = 1.875 * 10^-9
Input values into equation 1
S1 = 2.5885
ii) During 20-50
U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9
input values into equation 1
S2 = 1.5854 m
iii) During 50 -90
U3 = r^2*s / 4*π*t = 9.375 * 10^-10
input values into equation 1
S3 = 4.2888 m
iv) During 90-100
U4 = 0
s4 = 0
Drawdown at 100 days = ∑ Drawdowns at various period
= s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0
= 8.4627 m
Just because I seen someone else ask but they didn't have enough information.
If a filesystem has a block size of 4096 bytes, this means that a file comprised of only one byte will still use 4096 bytes of storage. A file made up of 4097 bytes will use 4096*2=8192 bytes of storage. Knowing this, can you fill in the gaps in the calculate_storage function below, which calculates the total number of bytes needed to store a file of a given size?
Answer:
Following are the program to the given question:
def calculate_storage(filesize):#definging a method calculate_storage that takes filesize as a parameter
block_size = 4096#definging block_size that holds value
full_blocks = filesize//block_size#definging full_blocks that divides the value and hold integer part
partial_block_remainder = filesize%block_size#definging partial_block_remainder that holds remainder value
if partial_block_remainder > 0:#definging if that compare the value
return block_size*full_blocks+block_size#return value
return block_size*full_blocks#return value
print(calculate_storage(1)) # calling method by passing value
print(calculate_storage(4096)) # calling method by passing value
print(calculate_storage(4097)) # calling method by passing value
Output:
4096
4096
8192
Explanation:
In this code, a method "calculate_storage" is declared that holds a value "filesize" in its parameters, inside the method "block_size" is declared that holds an integer value, and defines "full_blocks and partial_block_remainder" variable that holds the quotient and remainder value and use it to check its value and return its calculated value. Outside the method, three print method is declared that calls the method and prints its return value.
8. The operation of a TXV is controlled by the
O A. thermostatic spring.
O B. temperature bulb.
O C. external pressure of the evaporator.
O D. modulating valve.
For a sixth-order Butterworth high pass filter with cutoff frequency 3 rad/s, compute the following:
a. The locations of the poles.
b. The transfer function H(s).
c. The corresponding LCCDE description.
Solution :
Given :
A six order Butterworth high pass filter.
∴ n = 6, [tex]w_c=1 \ rad/s[/tex]
a). The location at poles :
[tex]$s^6-(w_c)^6=0$[/tex]
[tex]$s^6=(w_c)^6=1^6$[/tex]
∴ [tex]$s^6 = 1$[/tex]
Therefore, it has 6 repeated poles at s = 1.
b). The transfer function H(S) :
Transfer function H(S) [tex]$=\frac{1}{1+j\left(\frac{w_c}{s}\right)^6}$[/tex]
[tex]$=\frac{1}{1-\left(\frac{w_c}{s}\right)^6}$[/tex]
∴ H(S) [tex]$=\frac{s^6}{s^6-(w_c)^6}=\frac{s^6}{s^6-1}$[/tex]
H(S) [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]
c). The corresponding LCCDE description :
[tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]
[tex]$Y(s)(s^6-1) = s^6 \times (s)$[/tex]
[tex]$Y(s)s^6-y(s).1 = s^6 \times (s)$[/tex]
By taking inverse Laplace transformation on BS
[tex]$L^{-1}[Y(s)s^6-Y(s)1]=L^{-1}[s^6 \times (s)]$[/tex]
[tex]$\frac{d^6y(t)}{dt^6}-y(t)=\frac{d^6 \times (t)}{dt^6}$[/tex]
Hence solved.
A 14 bit A to D converter is to be employed with an Iron-constantan thermocouple. What temperature resolution can be expected with a full scale voltage of 100 mV? Thermocouple shows 0.5 mV at 50oC and 3.052 mV at 100oC.
Answer:
[tex]\gamma=0.122^oC[/tex]
Explanation:
From the question we are told that:
Word length[tex]n=14bit[/tex]
Full scale voltage of 100 mV
Thermo-couple shows:
[tex]0.5 mV\ at\ 50^oC[/tex]
[tex]3.052\ mV\ at\ 100^oC.[/tex]
Generally the equation for Average output voltage V_[avg} is mathematically given by
[tex]V_[avg}=\frac{3.052-0.5}{100-50}[/tex]
[tex]V_[avg}=0.05mv[/tex]
Since
[tex]Word lenght =14[/tex]
Therefore
[tex]Voltage\ steps=\frac{100}{2^{14}}[/tex]
[tex]Voltage\ steps=6.10*10^3mv steps[/tex]
Generally the equation for The required temperature resolution\gamma is mathematically given by
[tex]\gamma=\frac{100}{2^{14}*V_[avg}}[/tex]
[tex]\gamma=\frac{100}{2^{14}*0.05}[/tex]
[tex]\gamma=0.122^oC[/tex]
The AGC control voltage: ___________
a. varies as the signal strength of the received signal varies.
b. a negative feedback voltage.
c. is actually the dc voltage component produced by the mixing action in the AM demodulator stage.
d. is produced by an RC circuit having a much larger time constant than that of the detector.
e. all of the above
Answer:
The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.
Explanation:
Draw the logic circuit for each of the following. For each gate, determine if it generates either EVEN or ODD parity bit and finds the output for the given input data: (Remember: A XOR generates EVEN parity bit. A XNOR generates Odd Parity bit, whatever how many inputs they have.)
Data Inputs Which kind parity bit can it generate?
4-input XOR, input data-1001 Even Parity Bit ODD Parity Bit
5-input XOR, input data-10010
6-input XOR, input data-101001
7-input XOR, input data 1011011
Answer:
a) 4-input XOR, input data-1001 = 0 Even parity Bit
b) 5-input XOR, input data-10010 = 0 Even parity Bit
c) 6-input XOR, input data-101001 = 1 Even parity Bit
d) 7-input XOR, input data 1011011 = 1 Even parity Bit
Explanation:
a) 4-input XOR, input data-1001 ; generates 0 Even parity Bit
b) 5-input XOR, input data-10010 ; generates 0 Even parity Bit
c) 6-input XOR, input data-101001 ; generates 1 Even parity Bit
d) 7-input XOR, input data 1011011 ; generates 1 Even parity Bit
Attached below is the Logic circuits of the data inputs
Identify the best drying agent or process for each described purpose. Removal of small amounts of water from a polar solvent____. Removal of visible pockets of water from an organic solvent____. Storage of solvents or other materials in a desiccator_____.
Answer:
Calcium Chloride
Brine Wash
Drierite
Explanation:
Removal of small amounts of water from a polar solvent is Calcium Chloride
Removal of visible pockets of water from an organic solvent is Brine Wash
Storage of solvents or other materials in a desiccator is Drierite
In 2009 an explosive eruption covered the island of Hunga Ha'apai in black volcanic ash. What type of succession is this?
Answer:
The type of succession is:
Primary succession
Explanation:
This is a type of succession that occurs after a volcanic eruption or earthquake; it involves the breakdown of rocks by lichens to create new, nutrient rich soils.
Primary succession is one of the two types of succession we have. It begins on rock formations, such as volcanoes or mountains, or in a place with no organisms or soil.
An astronaut orbits the earth in a space capsule whose height above the earth is equal to the earth's radius. How does the mass of the astronaut in the capsule compare to her mass on the earth?
Answer:
The weight will be [tex]\frac{1}{4}[/tex] of its weight on earth surface. A further explanation is provided below.
Explanation:
According to the question,
h = R
The value of gravitational acceleration at height equivalent to radius of earth R.
⇒ [tex]g=\frac{g_0}{(1+\frac{h}{R} )^2}[/tex]
or,
⇒ [tex]g=\frac{g_0}{(1+\frac{R}{R} )^2} =\frac{g_0}{4}[/tex]
here,
[tex]g_0[/tex] = gravitational acceleration earth's surface
then,
⇒ [tex]mg=\frac{mg_0}{4}[/tex]
Thus, the above is the appropriate solution.
thiết kế ic 555 và code để ic hoạt động
Answer:
here you go.
screenshot 2 should give you some basic idea
When a voltage (v=353 sin (251t+30) is applied to two elements impedance a current (i =7.07 cos 251 t) is passing. Find the nature and the value of the elements and the circuit power
Answer:
A.C. voltage, V= V0 sin ωt As,t = πω = 12.2πω = 12T, therefore, first half cycle (T/2). Hence, average value of AC voltage, Eav = 2V0π.A sample of gage length 2 inches has received an elongation to 2.2 inches in tension and compressed to 1.818 inches in compression. The engineering stress in tension was 6 Mpsi and 6.2 Mpsi in compression. The area of cross section initially is 2 square inches.
Requried:
a. The true stress in tension is:_______
b. The true stress in compression is: ___________
c. The true strains in both tension and compression are : _________
Answer:
a. 6.6 Mpsi
b. 5.64 Mpsi
c. i. 0.095 ii. -0.095
Explanation:
a. The true stress in tension is:
The true stress in tension, σ = σ'(1 + ε) where σ' = engineering stress in tension = 6 Mpsi and ε = engineering strain under tension = (L₁ - L₀)/L₀ where L₀ = gage length = 2 inches and L₁ = length at elongation under tension = 2.2 inches. So, ε = (L₁ - L₀)/L₀ = (2.2 - 2)/2 = 0.2/2 = 0.1
So, σ = σ'(1 + ε)
σ = 6 Mpsi(1 + 0.1)
σ = 6 Mpsi(1.1)
σ = 6.6 Mpsi
b. The true stress in compression is:
The true stress in tension, σ₁ = σ"(1 + ε) where σ' = engineering stress in tension = 6 Mpsi and ε' = engineering strain under tension = (L₂ - L₀)/L₀ where L₀ = gage length = 2 inches and L₂ = length at elongation under compression = 1.818 inches. So, ε' = (L₂ - L₀)/L₀ = (1.818 - 2)/2 = -0.182/2 = -0.091
So, σ₁ = σ'(1 + ε')
σ₁ = 6.2 Mpsi[1 + (-0.091)]
σ₁ = 6.2 Mpsi[1 - 0.091)
σ₁ = 6.2 Mpsi(0.909)
σ₁ = 5.64 Mpsi
c. The true strains in both tension and compression are :
i. The true strain in tension
The true strain in tension ε" = ㏑(1 + ε) where ε = engineering strain in tension = 0.1
So, ε" = ㏑(1 + ε)
ε" = ㏑(1 + 0.1)
ε" = ㏑(1.1)
ε" = 0.095
ii. The true strain in compression
The true strain in tension ε₁ = ㏑(1 + ε') where 'ε = engineering strain in compression = -0.091
So, ε₁ = ㏑(1 + ε')
ε₁ = ㏑[1 + (-0.091)]
ε₁ = ㏑(1 - 0.091)
ε₁ = ㏑(0.909)
ε₁ = -0.095