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
Answer:
Use the right-hand rule for magnetic force to determine the charge on the moving particle.
This is a
negative
charge
Explanation:
Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus
ΔL = 1 m
Thus, the cable stretches by 1 m.
Answer:
k = 39.2 N / m
Explanation:
The 200 g block is accelerated by the force of friction between the blocks. Let's use Newton's second law
N- W = 0
N = W
fr = ma
μ N = ma
μ mg = ma
a=μ g
Let's look for the acceleration of the largest block that has oscillatory movement
x = A cos (w t)
A = 0.05 m
The maximum acceleration is cos wt = ±1
a = A w2
a = A k / m
We substitute and calculate
μ g = A k / M
k = μ g M / A
The mass that performs the oscillation is the mass of the two bodies
M = m1 + m2
k = 0.2 9.8 (0.800+ 0.200) /0.05
k = 39.2 N / m