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sladkih [1.3K]
2 years ago
5

State two advantages of sautoing​

Engineering
1 answer:
lbvjy [14]2 years ago
4 0
Sautéing is advantageous over certain other methods of cooking as it is a very fast process, and the amount of fat required is lesser as compared to deep-frying. However, sautéed dishes do contain a certain amount of fat, and are less healthy as compared to boiled or baked dishes.-google
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In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed
Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8  

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒  0.2\times 0.8

⇒  0.16

On the third attempt the success probability will be:

⇒  0.2\times 0.2\times 0.8

⇒  0.032

So that the success probability will be:

⇒  0.8 + 0.16 + 0.032

⇒  0.992

7 0
2 years ago
جائت فكرة ربط الحواسيب لغرض نقل البيانات و مشاركتها و بعدها بفتره قصيره جائت إمكانية مشاركة الموارد بين الحواسيب صح ام خطأ​
melamori03 [73]

Answer:

بدلاً من ذلك يُشار إليه باسم مشاركة أو مشاركة شبكة ، الدليل المشترك هو دليل أو مجلد يمكن الوصول إليه من قبل العديد من المستخدمين على الشبكة. هذه هي الطريقة الأكثر شيوعًا للوصول إلى المعلومات ومشاركتها على شبكة محلية

Explanation:

5 0
2 years ago
Read 2 more answers
What measures can be taken to mitigate or prevent the formation of stress corrosion cracks?
sasho [114]

Answer:

a) Selection and control of material

b) Control of stress

c) Control of the environment

Explanation:

The measures that can be taken to control and prevent the formation of strss corrosion cracks are:

a) Selection and control of material

It deals with the selection of the best quality material that can resist the stress corrosion cracks to a high extent.

b) Control of stress

It is to control or reduce the intensity of the major stress that causes the stress corrosion cracks.

c) Control of the environment

It includes the prevention of the material from the corrosive environment by removing the component from the region or controlling the environment condition around the object.

7 0
3 years ago
A reciprocating single-acting pump has a cylinder of 125mm bore and 300mm stroke and draws water from a sump whose water level i
djverab [1.8K]

Answer:

speed is 57.38 rpm

theoretical discharge =3.54 litres /min

Explanation:

given data

bore r = 125 mm

stock = 300 mm

water level = 1m

pipe L = 2 m

diameter = 50 mm

absolute pressure = 2.4 m

atmospheric pressure = 10.3 m

to find out

theoretical discharge and speed

solution

we will apply here pressure formula here

pressure (s) =  L/g × ( Area (p) / Area(s) ) × ω²×r×cos∅

for maximum pressure ∅ = 0

so put here all value

pressure (s) =  2/9.8 × ( π/4×125² / π/4×50² ) × ω²×300/2

pressure (s) = 0.191 ω²

and we know

atmospheric pressure = absolute pressure + pressure (s) + water level

10.3 = 2.4 + 1 + 0.191 ω²

solve it and we get

ω = 6 rad/sec

so discharge = A× L× N /60

so here N = 57.38

speed is 57.38 rpm

and

theoretical discharge = A(p) L N / 60

theoretical discharge = π/4×125² × 2 × 57.38 / 60

theoretical discharge =3.54 litres /min

5 0
2 years ago
Given the potential field in cylindrical coordinates, V = 100/ (z2+1) rho cos φV, and point P at rho = 3m, φ = 60°, z = 2m, find
Margaret [11]

Answer:

V = 30 V

vector (E) = -10*a_p + 17.32*a_Q - 24*a_z

mag(E) = 31.24 V/m

dV / dN = 31.2 V /m

a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z

p_v = 276 pC / m^3

Explanation:

Given:

- The Volt Potential in cylindrical coordinate system is given as:

                                     

- The point P is at p = 3 , Q = 60 , z = 2

Find:

values at P for:

a.) V

b.) E

c.) E

d.) dV/dN

e.) aN

f.) rhov in free space

Solution:

a)

Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:

                                     V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V      

b)

To compute the Electric field from Volt potential we have the following relation:

                                    E = - ∀.V

Where, ∀ is a del function which denoted:

                                   ∀ = \frac{d}{dp}.a_p + \frac{d}{p*dQ}*a_Q + \frac{d}{dz}*a_z

Hence,

                   E = \frac{100*cos(Q)}{z^2 + 1}.a_p+ \frac{100*sin(Q)}{z^2 + 1}.a_Q + \frac{-200*p*zcos(Q)}{(z^2 + 1)^2}.a_z

Plug in the values for point P:

                E = \frac{100*cos(60)}{5}.a_p+ \frac{100*sin(60)}{5}.a_Q + \frac{-200*3*2*cos(60)}{(5)^2}.a_z\\\\E = - 10*a_p +17.32 *a_Q-24*a_z

c)

The magnitude of the Electric Field is given by:

               E = √((-10)^2 + (17.32)^2 + (24)^2)

               E = √975.9824

               E = 31.241 N/C

d)

The dV/dN is the Field Strength E in normal to the surface with vector N given by:

              dV / dN = | - ∀.V |

              dV / dN = | E | = 31.2 N/C

e)

a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:

               a_N =  - vector(E) / (dV/dN)\\\\a_N =  10 /31.2 *a_p - 17.32/31.2 *a_Q + 24/31.2 *a_z\\\\a_N =  0.32 *a_p - 0.55 *a_Q + 0.77 *a_z

f)

The charge density in free space:

              p_v = E.∈_o

Where, ∈_o is the permittivity of free space = 8.85*10^-12

              p_v = 31.24.8.85*10^-12

              p_v = 276 pC / m^3

             

4 0
2 years ago
Read 2 more answers
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