A) In the case of the Boundary Thickness Layer we use the given formula,
We know as well that,
Re = Número de Reynolds = 
Where,
U = velocity
= kinematic viscosity
For water, kinematic viscosity, 
So, 
B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.
Thus,
C. Wall shear stress,
For water, dynamic viscosity, 
= 2.344*10^-5 lbf-s/ft^2
For two un-related quantities, the Heisenberg uncertainty equations holds: the prduct of the two uncertainty quantities is greater than
Example of unrelated quantities are position and momentum, energy and time.
Thus
Knowing the speed of the bacteria the uncertainty in its position is
Answer:
First Quarter and Third Quarter.
Explanation:
Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.
Since gravity variates with the distance:
(1)
Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.
For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.
When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).
However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.
Therefore, that happens when the Moon is at First Quarter and Third Quarter.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
