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choli [55]
2 years ago
13

The heat flux that is applied to the left face of a plane wall is q" = 20 w/m2. the wall is of thickness l = 10 mm and of therma

l conductivity k = 12 w/m · k. if the surface temperatures of the wall are measured to be 50°c on the left side and 30°c on the right side, do steady-state conditions exist?
Physics
2 answers:
inessss [21]2 years ago
5 0
Of course steady state condition occurs in almost any system but time it will occurs varies among system. for this kind of system, conduction, steady state conduction occurs when the temperature change from one point to the point is already constant. steady state is not achieved immediately because the heat travels and material will not be heated at the same way at the starting point.
ki77a [65]2 years ago
3 0

Answer:

given flux is very small then the above calculated value so it is not is steady state

Explanation:

As we know that heat flux is given in steady state by the formula

Q = \frac{k(T_1 - T_2)}{L}

here we know that

T_1 = 50^o C

T_2 = 30 ^o C

k = 12 W/m k

L = 10 mm

now we have

Q = \frac{12(50 - 30)}{0.01}

Q = 24000 w/m^2

since given flux is very small then the above calculated value so it is not is steady state

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Ludmilka [50]

Answer:

D. 1

Explanation:

I believe this is the answer

6 0
1 year ago
Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot
Mnenie [13.5K]

Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

     F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

    q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

    x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

    q₂ = -6.0 nC = -6.0 10⁻⁹ C

    x₂ = 4.0 cm = 4.0 10⁻² m

    q₃ = 5.0 nC = 5.0 10⁻⁹ C

    x3 = 0 m

Charges q1 and q3

    r = x₁ -x₃

    r = 2.0 10⁻² -0

    r = 2.0 10⁻² m

    F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

    F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

    r = r₂ –r₃

    r = 4.0 10⁻² - 0 = 4.0 10⁻² m

    F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

    F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

    F = F₂₃ - F₁₃

    F = 16.86 10⁻⁵ - 33.7 10⁻⁵

    F = - 16.84 10⁻⁵ N

    F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis

5 0
3 years ago
Ma puteti ajuta sa rezolv la fizica o problema ?
xxTIMURxx [149]
Care este problema? Btw why are you speaking Romanian 
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3 years ago
3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one
hoa [83]

Here are the observations

<u>S</u><u>u</u><u>g</u><u>a</u><u>r</u><u>:</u><u>-</u>

  • Sugar is soluble in water
  • so It will dissolve in water .

<u>C</u><u>o</u><u>r</u><u>n</u><u> </u><u>s</u><u>y</u><u>r</u><u>u</u><u>p</u><u>:</u><u>-</u>

  • Corn syrup is also basically a sugar.
  • It will dissolve in water too .
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<u>O</u><u>i</u><u>l</u><u>:</u><u>-</u>

  • Oil is not soluble in water
  • Hence it won't dissolve in water.
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2 years ago
Question 14 of 25
natali 33 [55]

the answer is sueist

Explanation:

7 0
2 years ago
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