Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
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Answer:
μ = 0.309
Explanation:
coefficient of kinetic friction is defined as the ratio of two forces, friction force and the normal force acting on the object.
θ = arctan(15/100)= 8.531⁰
In the vertical direction:
N = mgcosθ = 100 *9.8 *cos(8.531) = 970N
law of conservation of energy implies
mgsinθ - μNx = 1/2m(v₂²-v₁²)
100*9.8*sin (8.531) - μ(970*2) = 1/2(100)(0²-3²)
150.6 - 1940μ = 450
- 1940μ = -600.6
μ = 0.309
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Explanation:
