Consider the following steady, two-dimensional velocity field: V→(u, v) = (0.46 + 2.1x)i→ + (−2.8 - 2.1y)j→ Calculate the location of the stagnation point. The location of the stagnation points are x = and y = .

there is no stagnation point

Step-by-step explanation:

for the velocity field V→(u,v)= (0.46 + 2.1x)i→ + (−2.8 - 2.1y)j , the stagnation point is found when the velocity vectors converge in one point ( thus also stays in that place when the point is reached). Thus the stagnation point can be found when the divergence of the velocity field is <0 ( thus it does not diverge , but converges)

div(V) = ∇*V= d/dx (0.46 + 2.1x) + d/dy (−2.8 - 2.1y) = 0

2.1 - 2.1 = 0

since div(V) can never be  <0 , there is no stagnation point

the radius is 2 times the diameter.

step-by-step explanation: