We are given with
L = 0.483 m
s = 1.23 cm = 0.0123 m (side of square)
T1 = 20.0 C
T2 = 80.8 C
This is a conductive heat transfer problem. We use this formula:
q = k A L (T2 - T1)
Since the rod is made out of two materials, the formula will be
q = ki Ai L (T2 - T1) + ks As L (T2 - T1)
where the subscript i refers to iron
and s to stainless steel
From literature:
ki = 55 W/ m K
ks = 29 W/m K
Using geometry:
As = 1.23²
Ai = π(1.23√2)²/4 - 1.23²
Substitute the values into the formula and solve for q<span />
Answer:
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Answer:
4 %
Explanation:
mass of the object 1, m₁ = 0.1 Kg
initial velocity of object 1,v₁= 0.2 m/s
mass of object 2, m₂ = 0.15 kg
initial speed of the object 2. v₂ = 0 m/s
Kinetic energy retained by the object 1 = ?
using the formula to calculate velocity of the ball after collision
negative sign represent the velocity is in opposite direction after collision.
Initial KE
final KE
% of KE retained
=
=
= 4 %
the percentage of Kinetic energy retained after collision is equal to 4 %