The shells correspond with the principal quantum numbers (n<span> = </span>1<span>, </span>2<span>, </span>3<span>, 4 ...) Each shell can contain only a fixed number </span>of electrons. The first shell can hold up to two electrons<span>, the second shell can hold up to eight </span>electrons<span>, the third shell can hold up to 18 and so on.</span>
Answer:The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to m, the mass of the oscillating object. The speed of sound in air is low, because air is easily compressible.
Explanation:
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
Explanation:
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / F
F = 50 / cos 28°
F = 56.6 lbs
