quy trình sản xuất bao bì plastic dạng túi

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

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:

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:

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.

Explanation:

sory sorry sorry sorrysorrysorry

Answer:

25 mins into hours = 0.416667 hours

25 inches as mm = 635

Explanation:

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

Answer:

a. True

b. False

Explanation:

a. Since true stress, σ' = σ(1 + ε) where σ = engineering stress and ε = engineering strain.

Also under tension ε > 0, so, (1 + ε) > 1

Since (1 + ε) = σ'/σ > 1, ⇒ σ' > σ

So, the true stress is greater than the engineering stress.

So, the statement is true

b. Creep is a time-dependent deformation under certain applied load.

Creep occurs at high temperatures under different types of stress.

But, Creep test is carried out at constant high temperature and constant stress.

This statement is false.

Answer:

364 566 inches of class-6th from a

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

Calculate the kinematic viscosity of Fluid A

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

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:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

Calculate the change in the enthalpy of R-134a

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant  

hence ; s1 = s2

by interpolation  ; h2 = 106.95

Δ enthalpy = h2 - h1

                  =  ( 106.95 - 129.85 ) = -23 Btu/Ibm

Answer:

5354

Explanation:

Answer:

Option C = Only 1 path for current to flow.

Answer:

For small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed. qf = final temperature of object

Explanation:

hope this helps you sorry if it doesn’t help you

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:

here you go.

screenshot 2 should give you some basic idea

Answer:

bood ekogcd gcaerh is an American fbnuxc

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] (%)  

Answer:

13.86 MPa

Explanation:

We need to find the change in length ΔL of the steel beam and thus, the strain in the beam, ε.

ΔL = L₀αΔθ

Strain, ε = ΔL/L₀ = αΔθ where α = coefficient of linear expansion of steel = 11 × 10⁻⁶ °C and Δθ = temperature change of steel beam = θ₂ - θ₁ where θ₁ = initial temperature of steel beam = 23 °C and θ₂ = final temperature of steel beam = 42 °C

Substituting the values of the variables into the equation, we have

ε = αΔθ

ε = α(θ₂ - θ₁)

ε = 11 × 10⁻⁶ /°C(42 °C  - 23 °C )

ε = 11 × 10⁻⁶ /°C(42 °C)

ε = 462 × 10⁻⁶

ε = 4.62 × 10⁻⁴

Now, the Young's modulus of concrete is Y = 30 GPa = 30 × 10⁹ N/m²

Y = σ/ε where σ = compressional stress from concrete and ε = strain on steel

So, σ = Yε

σ = 30 × 10⁹ N/m² × 4.62 × 10⁻⁴

σ = 138.6 × 10⁵ N/m²

σ = 1.386 × 10² × 10⁵ N/m²

σ = 1.386 × 10⁷ N/m²

σ = 13.86 × 10⁶ N/m²

σ = 13.86 MPa

Answer:

The basic difference between Total cost and total revenue is that the total cost includes the total expenditure incurred on the production of a commodity whereas total revenue refers to the money received from selling that commodity.

Explanation:

Answer:

its A

Explanation:

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:

60000 H

Explanation:

We are given;

Number of turns; N = 3000 turns

Change in flux = 900 - 860 = 40 Wb

Change in current = 9 - 7 = 2 A

Now, the formula for incremental inductance is given as:

L = N(Change in flux/Change in current) where;

N is Number of turns

Plugging in the relevant values, we have;

L = 3000(40/2)

L = 60000 H.

Answer:

soil>>Slop sink

>Urinal

>water closet

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:

90

Explanation:

mean is basically taking the sum of all numbers and then dividing the sum with the number of all given numbers..

here, the mean is 9, total numbers are 10.. so the sum will be 9 multiplied by 10, that is 90.

If the mean of 10 numbers is 9, then the sum (total) of these numbers will be​ 90. The correct option is D.

The mean is the average of a set of variables in mathematics and statistics. The mean can be calculated in a variety of ways, including the simple arithmetic mean (add the numbers and divide the total by the number of observations), geometric mean, and harmonic mean. Mean is denoted by a small m.

To calculate mean: Arrange the data items in ascending order from least to greatest. If the number of points is odd, the median is the data point in the middle of the list. If the number of data points in the list is even, the median is the average of two middle data points.

Mean = total sum of the numbers/total numbers

9 = total sum of the numbers / 10

The sum of the terms = 9 x 10 = 90

90 = total sum of the numbers

Therefore, the correct option is D. 90

To learn more about the mean, refer to the link:

https://brainly.com/question/28670966

#SPJ2

The question is incomplete. Your most probably complete question is given below:

A. 9. B. 0.9. C. 70. D. 90

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.