<span>Health problems that develop later</span>
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:
104.3 cm or 179.7
Explanation:
First find time that it takes for the object to hit the ground
*
Then find xf of projectile 
not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7
hope that helps
Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding,
= 50 turns
number of turns in the secondary winding,
= 10 turns
the secondary load resistance,
= 250 Ω
Determine the turns ratio;

Now, determine the reflected resistance in the primary winding;

Therefore, the reflected resistance in the primary winding is 6250 Ω
Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...