The temperature within a thin plate with thermal conductivity of 10 W/m/K depends on position as given by the following expressi
on: TT=(100 K)????????−xx2/????????xx 2cos�yy/????????yy�+300 K Where, Lx = 1 m, and Ly = 2 m. At the point (0.4 m, 1 m), find: a. The magnitude of the heat flux b. The direction of the heat flux
1 answer:
Answer:
Heat flux = (598.3î + 204.3j) W/m²
a) Magnitude of the heat flux = 632.22 W/m²
b) Direction of the heat flux = 18.85°
Explanation:
- The correct question is the first image attached to this solution.
- The solution to this question is contained on the second and third images attached to this solution respectively.
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Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 × m
separation L = 0.5 mm = 0.5 × m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 × m
λ = 2042 nm
Answer:
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Explanation:
Answer:
θ = 33.8
a = 3.42 m/s²
Explanation:
given data
mass m = 2 kg
coefficient of static friction μs = 0.67
coefficient of kinetic friction μk = 0.25
solution
when block start slide
N = mg cosθ .............1
fs = mg sinθ ...............2
now we divide equation 2 by equation 1 we get
= tanθ
put here value we get
tan θ = 0.67
θ = 33.8
and
when block will slide then we apply newton 2nd law
mg sinθ - fk = ma ...............3
here fk = μk N = μk mg cosθ
so from equation 3 we get
mg sinθ - μk mg cosθ = ma
so a will be
a = (sinθ - μk cosθ)g
put here value and we get
a = (sin33.8 - 0.25 cos33.8) 9.8
a = 3.42 m/s²
