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Devinda W.
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Physics (Heat Transfer)
TutorMe
Question:

Chromium steel ball bearing (π‘˜ = 50𝑀/π‘šπΎ, 𝛼 = 1.3 Γ— 10βˆ’5π‘š^2/𝑠) are to be heat treated. They are heated to a temperature of 650 𝐢 and then quenched in oil that is at a temperature of 55 𝐢. The ball bearings have a diameter of 4cm and the convective heat transfer coefficient between the bearings and oil is 300𝑀/π‘š2𝐾. Determine the length of time that the bearings must remain in oil before the temperature drops to 200 𝐢.

Devinda W.
Answer:

Chromium steel ball bearings π‘˜ = 50𝑀/π‘šπΎ 𝛼 = π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™ 𝑑𝑖𝑓𝑓𝑒𝑠𝑖𝑣𝑖𝑑𝑦 = 1.3 Γ— 10βˆ’5π‘š^2/𝑠 π‘‡π‘œ = 650 𝐢, π‘‡βˆž = 55 𝐢 Diameter of ball bearing, 𝑑 = 4π‘π‘š = 0.04π‘š β„Ž = 300𝑀/π‘š2𝐾 𝜏 = ? 𝑇(𝑑) = 200 𝐢 𝐡𝑖 =β„Ž 𝐿/π‘˜ Characteristic length, 𝐿 = volume of the ball/surface area of the ball = 𝑉/𝐴 ∴ 𝐿 = 𝑣/𝐴 = (4πœ‹π‘Ÿ^3 /3) / 4πœ‹π‘Ÿ^2 = π‘Ÿ/3 𝐡𝑖 = β„Ž 𝐿/π‘˜ = (300 Γ— 0.04) / (6 Γ— 50) = 0.04 Since 𝐡𝑖 β‰ͺ 0.1, the lumped capacity system or the negligible internal resistance the theory is valid. πœƒ/πœƒπ‘œ = (𝑇(𝑑) βˆ’ π‘‡βˆž) / (π‘‡π‘œ βˆ’ π‘‡βˆž) = 𝑒^ (βˆ’π΅π‘– πΉπ‘œ) πΉπ‘œ = π‘˜ *𝜏/ (πœŒπ‘π‘ƒπΏ^2) = 𝛼*𝜏 / 𝐿^2 = 1.3 Γ— 10βˆ’5 / (0.02/3)^2 = 0.2925 𝜏 πœƒ/πœƒπ‘œ = (200 βˆ’ 55) / 650 βˆ’ 55) = 𝑒 ^(βˆ’0.04Γ—0.2925 𝜏) 0.2437 = 𝑒^(βˆ’0.0117 𝜏) log 0.2437 = log 𝑒^(βˆ’0.0117 𝜏) = βˆ’0.0117 𝜏 log 𝑒 ∴ 𝜏 = log 0.2437/ (log 𝑒 Γ— βˆ’0.0117 ) = log 0.2437/ βˆ’0.0117 log 𝑒 = 120.7 𝑠

Trigonometry
TutorMe
Question:

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

Devinda W.
Answer:

sin(A) = 1βˆ’sin(A)^2 sin(A) = cos(A)^2 sin(A)^2 = cos(A)^4 1βˆ’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

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).

Devinda W.
Answer:

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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