Answer:
The temperature of the Aluminium plate 44.84⁰C
Explanation:
Number of transistors = 4
Since the heat dissipated by each transistor is 12W
Total heat dissipated, Q = 4 * 12 = 48 W
Q = 48 W
Cross sectional Area of the Aluminium plate, A = 2(l * b)
l = Length of the aluminium plate = 22 cm = 0.22 m
b = width of the aluminium plate = 22 cm = 0.22 m
A =2( 0.22 * 0.22 )
A = 0.0968 m²
From the heat balance equation, Q = hAΔT
h = 25 W/m²·K
A = 0.0968 m²
ΔT = T - T(air)
T(air) = 25°C
ΔT = T - 25°C
Q = 25 * 0.0968 * ( T - 25)
Q = 2.42 (T - 25)
Substitute Q = 48 into the equation above
48 = 2.42 (T - 25)
T - 25 = 19.84
T = 25 + 19.84
T = 44.84 ⁰C
Answer:
2.5 * 10^-3
Explanation:
<u>solution:</u>
The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:
<em>δu/δx+δv/δy=0</em>
so that:
<em>δv/δy= -δu/δx</em>
Now, since u = Uy/δ, where δ = cx^1/2, we have that:
<em>u=U*y/cx^1/2</em>
and we obtain:
<em>δv/δy=U*y/2cx^3/2</em>
The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):
v=∫δv/δy(dy)=U*y/4cx^1/2
=y/x*(U*y/4cx^1/2)
=u*y/4x
which is exactly what we needed to demonstrate.
Also, using u = U*y/δ in the last equation we can obtain:
v/U=u*y/4*U*x
=y^2/4*δ*x
which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:
(v/U)_max=δ^2/4δx
=δ/4x
=2.5 * 10^-3
Answer:
98 joules
Explanation:
1/2 of 28 = 14
14 multiply by 7= 98 joules
Answer: Yes the further the sun is away the longer the shadow is. At noon,the shadow is the shortest because its straight up above you. If this helps pls mark brainliest!
