Answer:
We are given the trajectory of a projectile:
y=H+xtan(θ)−g2u2x2(1+tan2(θ)),
where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then
H+xtan(θ)−12Lx2(1+tan2(θ))=0
Differentiate both sides with respect to θ.
dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0
Solving for dxdθ yields
dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))
This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.
H+L−12Lx2−L2=0.
This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.
Answer:

Explanation:
An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

The power consumed by the compressor can be calculated by the following expression:

Let consider that air behaves ideally. The density of air at inlet is:





The mass flow through compressor is:



The work input is:


Answer:
The direction will be NE i.e North east.
Explanation:
From the question, it will be travelling in North east direction because the the negative X components in vector is in horizontal direction which is the Eastern direction when using the four cardinal points and the Y components, the positive is in North direction if you use the four cardinal points. Therefore, the vector with both X components and Y components will be travelling in Northeast direction.
Answer:
s (l = 0) only two electrons
p (l = 1) sublevel six (6) electrons fit
d (l = 2) sub-level has 10 electrons
f (l = 3) has 7 has 14 electrons
Explanation:
The electrons have an odd spin, so they must comply with the Pauli principle, therefore only two electrons fit in a given sublevel.
Sub level s (l = 0) only two electrons
Sub level p (l = 1) has three orbitals in which two electrons fit, therefore in the sublevel six (6) electrons fit
Sub-level d (l = 2) has 5 orbitals each with two electrons, the sub-level has 10 electors
Sub-level f (l = 3) has 7 orbitals and two electrons each, the sub-level has 14 electrons