# Tutor profile: Devinda W.

## Questions

### Subject: Physics (Heat Transfer)

Chromium steel ball bearing (π = 50π€/ππΎ, πΌ = 1.3 Γ 10β5π^2/π ) are to be heat treated. They are heated to a temperature of 650 πΆ and then quenched in oil that is at a temperature of 55 πΆ. The ball bearings have a diameter of 4cm and the convective heat transfer coefficient between the bearings and oil is 300π€/π2πΎ. Determine the length of time that the bearings must remain in oil before the temperature drops to 200 πΆ.

Chromium steel ball bearings π = 50π€/ππΎ πΌ = π‘βπππππ πππππ’π ππ£ππ‘π¦ = 1.3 Γ 10β5π^2/π ππ = 650 πΆ, πβ = 55 πΆ Diameter of ball bearing, π = 4ππ = 0.04π β = 300π€/π2πΎ π = ? π(π‘) = 200 πΆ π΅π =β πΏ/π Characteristic length, πΏ = volume of the ball/surface area of the ball = π/π΄ β΄ πΏ = π£/π΄ = (4ππ^3 /3) / 4ππ^2 = π/3 π΅π = β πΏ/π = (300 Γ 0.04) / (6 Γ 50) = 0.04 Since π΅π βͺ 0.1, the lumped capacity system or the negligible internal resistance the theory is valid. π/ππ = (π(π‘) β πβ) / (ππ β πβ) = π^ (βπ΅π πΉπ) πΉπ = π *π/ (ππππΏ^2) = πΌ*π / πΏ^2 = 1.3 Γ 10β5 / (0.02/3)^2 = 0.2925 π π/ππ = (200 β 55) / 650 β 55) = π ^(β0.04Γ0.2925 π) 0.2437 = π^(β0.0117 π) log 0.2437 = log π^(β0.0117 π) = β0.0117 π log π β΄ π = log 0.2437/ (log π Γ β0.0117 ) = log 0.2437/ β0.0117 log π = 120.7 π

### Subject: Trigonometry

If sin(A)+sin(A)^2 = 1 and a*cos(A)^12+b*cos(A)^8+c*cos(A)^6β1=0 then b+c/a+b=?

sin(A) = 1βsin(A)^2 sin(A) = cos(A)^2 sin(A)^2 = cos(A)^4 1βcos(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

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).

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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