To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

Where:
are the viscosities of the concrete before and after the increase
l = Length of the vessel
= Radio of the vessel before and after the increase
= Change in the pressure
The rates of flow before and after he increase
Our values are given as:
10 times her resting rate
95% of its normal value
Increase of 50%
Plugging known information to get







Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.
The wire needs to be sauderwired to be connected back into place to get energy into column so it came function properly again!
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.
The Celsius degree is the same size as the Kelvin.
The correct choice is 'C'.
I believe it is speed.
Hope this helps!