Answer:
The average gauge pressure inside the vein is 110270.58 Pa
Explanation:
This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

I believe the answer is Mouse/Herbivore. It was a question in my old lab haha.
Hope this helped!!
~xox Melis
Answer:
<h3>Because one Coulomb of charge is an abnormally large quantity of charge, the units of microCoulombs (µC) or nanoCoulombs (nC) are more commonly used as the unit of measurement of charge. To illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to have a total charge of -1 C.</h3>
Explanation:
<h3><em><u>mark as brainliast</u></em></h3><h3><em><u>indian </u></em><em><u>genius </u></em><em><u>s</u></em><em><u>a</u></em><em><u>r</u></em><em><u>thak</u></em></h3>