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s344n2d4d5 [400]

3 years ago

11

A Newtonian fluid with a viscosity  drains through the space between two large parallel plates as shown in the figure. The gap

distance between the plates is 2b. Obtain relations for the shear stress distribution, shear stress at the walls, velocity profile and volumetric flow rate assuming laminar flow and negligible end effects. You can begin the derivations by considering that vz=vz(x) only and pressure at the inlet and outlet is atmospheric.

1 answer:

Rina8888 [55]3 years ago

8 0

Answer:

well

Explanation:

basically it's like

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Barnard's star is a near neighbor of the Sun whose properties we know quite well. It is a type M4V with absolute magnitude 13.22

vitfil [10]

Answer:

The star is at a distance of 100 parsecs.

Explanation:

The distance can be determined by means of the distance modulus:

$M - m = 5log(d) - 5$ (1)

Where M is the absolute magnitude, m is the apparent magnitude and d is the distance in units of parsec.

Therefore, d can be isolated from equation 1

$log(d) = (M - m + 5)/5$

Then, Applying logarithmic properties it is gotten:

$d = 10^{(M - m + 5)/5}$ (2)

The absolute magnitude is the intrinsic brightness of a star, while the apparent magnitude is the apparent brightness that a star will appear to have as is seen from the Earth.

Since both have the same spectral type is absolute magnitude will be the same.

Finally, equation 2 can be used:

$d = 10^{(13.22 - 8.22+ 5)/5}$

$d = 100 pc$

Hence, the star is at a distance of 100 parsecs.

Key term:

Parsec: Parallax of arc seconds

8 0

3 years ago

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved

Alina [70]

Answer:

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let $d$ represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

the line joining the microphone and the center of the two speakers, and
the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point $P_0$ would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$ .

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$ .

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$ .

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$ .

Calculate $\theta$ for each of these path differences:

$\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}$ .

In each of these case, the distance between the microphone and $P_0$ would be $9.4\, \tan(\theta)$ . Therefore:

At the first minimum, the distance from $P_0$ is approximately $1.1\; \rm m$ .
At the first maximum, the distance from $P_0$ is approximately $2.2\; \rm m$ .
At the second minimum, the distance from $P_0$ is approximately $3.4\; \rm m$ .
At the second maximum, the distance from $P_0$ is approximately $4.7\;\rm m$ .

6 0

4 years ago

Two material objects cannot occupy the same space at the same time, but two different disturbances of matter can. True or False?

Alex777 [14]

Answer:

false

Explanation:

3 0

4 years ago

Read 2 more answers

Cathode rays were shown to be a stream of _____.

guapka [62]

Cathode rays were shown to be a stream of "electrons".

Cathode rays (likewise called an electron beam) are streams of electrons saw in vacuum tubes. In the event that a cleared glass tube is outfitted with two anodes and a voltage is connected, the glass inverse the negative terminal is seen to sparkle from electrons radiated from the cathode. Electrons were first found as the constituents of cathode beams. The picture in an exemplary TV is made by centered light emission redirected by electric or magnetic fields in cathode ray tubes (CRTs).

7 0

4 years ago

Read 2 more answers

Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri

Romashka-Z-Leto [24]

The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

$E = \frac{Q}{4\pi \varepsilon_o r^2}$

where;

E is the electric field
Q is the charge
r is the radius

The electric field reduces by a factor of $\frac{1}{r^2}$

<h3>Surface area of a Gaussian surface;</h3>

The surface area of a sphere is given as;

$A = 4\pi r^2$

<h3>Change in area with r</h3>

$\frac{dA}{dr} = 8\pi r$

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446

7 0

2 years ago