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_____.

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.

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]

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.

Answer:

The correct solution is "5.74%".

Explanation:

The given values are:

Gravity of aggregate,

[tex]G_{agg}=2.7[/tex]

Gravity of asphalt,

[tex]G_{asp}=1.0[/tex]

Asphalt concrete mixture,

[tex]W_{agg}=0.94 \ W_m[/tex]

We know that,

[tex]W_{asp}=W_m-W_{agg}[/tex]

        [tex]=0.06 \ W_m[/tex]

Now,

The theoretical specific gravity of mix,

⇒ [tex]G_t=\frac{W_{agg}+W_{asp}}{\frac{W_{agg}}{G_{agg}} +\frac{W_{asp}}{G_{asp}} }[/tex]

By putting the values, we get

         [tex]=\frac{0.94 \ Wm+0.06 \ Wm}{\frac{0.94 \ Wm}{2.7} +\frac{0.06 \ Wm}{1} }[/tex]

         [tex]=2.45[/tex]

hence,

The percentage of voids will be:

⇒  %V = [tex]\frac{G_t-G_m}{G_t}\times 100[/tex]

           = [tex]\frac{2.45-2.317}{2.45}\times 100[/tex]

           = [tex]\frac{0.133}{2.317}\times 100[/tex]

           = [tex]5.74[/tex] (%)  

Explanation:

sory sorry sorry sorrysorrysorry

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

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

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.

Answer:

For  lower disk :   V = e^θ​rω(0)/h  = 0

At the upper disk:  V = e^θ​rω(h)/h  = e^θ​rω

Hence The physical boundary conditions are satisfied

Explanation:

Velocity field ( V ) = e^θ​rωz/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^θ​rω(0)/h  = 0

At the upper disk  V = e^θ​rω  given that a upper disk it rotates at z = h

∴ V = e^θ​rω(h)/h  = e^θ​rω

Hence we can conclude that the velocity field satisfies the appropriate physical boundary conditions.

Answer:

soil>>Slop sink

>Urinal

>water closet

Answer:

364 566 inches of class-6th from a

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]

Answer:

Nguyên lý hoạt động của cảm biến từ

Cảm biến từ có nguyên lý hoạt động khá đơn giản. Theo đó, một cuộn cảm sẽ phát triển một từ trường khi có dòng điện chạy qua nó; hay nói cách khác một dòng điện sẽ chạy qua một mạch chứa cuộn cảm khi từ trường xuyên qua nó thay đổi.

Ngoài ra, hiệu ứng này còn có thể được sử dụng để phát hiện các vật kim loại có tương tác với từ trường. Các chất phi kim loại như chất lỏng hay một số loại bụi bẩn không tương tác với từ trường. Do đó, cảm biến từ hoàn toàn có thể hoạt động được trong điều kiện ẩm ướt hoặc bụi bẩn.

Explanation:

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

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

Answer:

25 mins into hours = 0.416667 hours

25 inches as mm = 635

Explanation:

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.

1. Property Animation

2. View Animation

3. Drawable Animation

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.

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.

Answer:

The answer is below

Explanation:

a) The work done is equal to the loss in kinetic energy (KE)

Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy

Final KE = [tex]\frac{1}{2}mv_f^2[/tex]

But the final velocity is 0 (at rest). Hence:

Final KE = [tex]\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0[/tex]

ΔKE = 0 - K = -K

W = ΔKE = -K

Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)

W = qEd

but q = -e, hence:

W = -e * E * d

Using:

W = ΔKE

-e * E * d = -K

E= K / (e * d)

b) The electric field is in the direction of the electrons motion

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;

M, O, U or O, M, U.

- 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

U, N, T or T, N, U

Read more about Logical reasoning at; https://brainly.com/question/14458200

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.

False, Good housekeeping eliminates accident and fire hazards. It also maintains safe, healthy work conditions; saves time, money, materials, space, and effort; improves productivity and quality; boosts morale; and reflects an image of a well-run, successful organization.

Hope it helps you❤️

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]

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

Answer:

Answer:

here you go.

screenshot 2 should give you some basic idea

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!