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# Tutor profile: Devinda W.

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Devinda W.
Tutor for five years
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## Questions

### Subject:Physics (Heat Transfer)

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Question:

Chromium steel ball bearing (𝑘 = 50𝑤/𝑚𝐾, 𝛼 = 1.3 × 10−5𝑚^2/𝑠) are to be heat treated. They are heated to a temperature of 650 𝐶 and then quenched in oil that is at a temperature of 55 𝐶. The ball bearings have a diameter of 4cm and the convective heat transfer coefficient between the bearings and oil is 300𝑤/𝑚2𝐾. Determine the length of time that the bearings must remain in oil before the temperature drops to 200 𝐶.

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Devinda W.

Chromium steel ball bearings 𝑘 = 50𝑤/𝑚𝐾 𝛼 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 = 1.3 × 10−5𝑚^2/𝑠 𝑇𝑜 = 650 𝐶, 𝑇∞ = 55 𝐶 Diameter of ball bearing, 𝑑 = 4𝑐𝑚 = 0.04𝑚 ℎ = 300𝑤/𝑚2𝐾 𝜏 = ? 𝑇(𝑡) = 200 𝐶 𝐵𝑖 =ℎ 𝐿/𝑘 Characteristic length, 𝐿 = volume of the ball/surface area of the ball = 𝑉/𝐴 ∴ 𝐿 = 𝑣/𝐴 = (4𝜋𝑟^3 /3) / 4𝜋𝑟^2 = 𝑟/3 𝐵𝑖 = ℎ 𝐿/𝑘 = (300 × 0.04) / (6 × 50) = 0.04 Since 𝐵𝑖 ≪ 0.1, the lumped capacity system or the negligible internal resistance the theory is valid. 𝜃/𝜃𝑜 = (𝑇(𝑡) − 𝑇∞) / (𝑇𝑜 − 𝑇∞) = 𝑒^ (−𝐵𝑖 𝐹𝑜) 𝐹𝑜 = 𝑘 *𝜏/ (𝜌𝑐𝑃𝐿^2) = 𝛼*𝜏 / 𝐿^2 = 1.3 × 10−5 / (0.02/3)^2 = 0.2925 𝜏 𝜃/𝜃𝑜 = (200 − 55) / 650 − 55) = 𝑒 ^(−0.04×0.2925 𝜏) 0.2437 = 𝑒^(−0.0117 𝜏) log 0.2437 = log 𝑒^(−0.0117 𝜏) = −0.0117 𝜏 log 𝑒 ∴ 𝜏 = log 0.2437/ (log 𝑒 × −0.0117 ) = log 0.2437/ −0.0117 log 𝑒 = 120.7 𝑠

### Subject:Trigonometry

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Question:

If sin(A)+sin(A)^2 = 1 and a*cos(A)^12+b*cos(A)^8+c*cos(A)^6−1=0 then b+c/a+b=?

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Devinda W.

sin(A) = 1−sin(A)^2 sin(A) = cos(A)^2 sin(A)^2 = cos(A)^4 1−cos(A)^2 = cos(A)^4 1 = cos(A)^4 + cos(A)^2 1^3 = (cos(A)^4 + cos(A)^2 ) ^3 by formula (a+b)^3 1 = cos(A)^12 +3*cos(A)^10 + 3*cos(A)^8 + cos(A)^6 Transverse 1 onto the other side we get the condition as given as in the question. Comparing the variables we get a=1, b=3, c=1, hence the value of b+c/a+b=3+1/1 + 3 = 7

### Subject:Algebra

TutorMe
Question:

f(x) is a quadratic function such that f(2) =6 and f(4) = 6. Find the x coordinate of the vertex of the graph of f(x).

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Devinda W.

A. f(x) = ax2 + bx + c B. f(2) = 6 which give 6 = 4a + 2b + c C. f(4) = 6 which gives 6 = 16a + 4b + c D. 12a + 2b = 0 : subtract equation B from equation C Differentiate equation A from x, gives, d(f(x))/d(x) = 2ax + b At min/max point d(f(x))/d(x) = 0 . therefore x = -b/2a From equation D, -b/2a = 3. Therefore x = 3

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