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
Considering the deuterium-tritium fusion reaction with the tritium nucleus at rest: ¹₂H + ¹₃H → ²₄He + ⁰₁n the electric potential energy (in electron volts) at this distance is 17.58MeV
<h3>How is the electric potential energy of deuterium-tritium fusion reaction calculated?</h3>
The reaction is ¹₂H + 1₃H → ²₄He + ⁰₁n
Value of Q = (Mass of ¹₂H + Mass of ¹₃H - Mass of ²₄He- Mass of n) x 931 MeV
Mass of ¹₂H = 2.014102
Mass of ¹₃H = 3.016049
Mass of ²₄He = 4.002603
Mass of n = 1.00867
Therefore Value of Q = [2.014102+3.016049−4.002603−1.00867] × 931 MeV
Therefore Value of Q = 0.01887 × 931 MeV
= 17.58MeV
To learn more about deuterium-tritium fusion reaction, refer
brainly.com/question/9054784
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Answer:
Wave A
<em>I</em><em> </em><em>hope this</em><em> </em><em>helps</em><em> </em>
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
The answer is 4.0075 x 10^9
