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3 years ago

A(n) ______ is used to measure fluid flow in engineering

Engineering

1 answer:

Kazeer [188]3 years ago

6 0

Explanation:

instrument of engineering and management skills that can be used to identify a device on a network of devices

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What is the fastest motorcycle in the world ?

givi [52]

Answer:

Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...

MTT Turbine Superbike Y2K – top speed: 227 mph. This bike is one of the most powerful production motorcycles. ...

Suzuki Hayabusa – top speed: 248 mph. 1340cc

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3 years ago

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Can u say what’s this

tatuchka [14]

Answer:

particles of a solid object packed together

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3 years ago

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Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

3. In the text, we described a multithreaded file server, showing why it is better than a single-threaded server and a finite-st

NemiM [27]

Answer and Explanation:

• 1 thread awaits the incoming request

• 1 thread responds to the request

• 1 thread reads the hard disk

A multithreaded file server is better than a single-threaded server and a finite-state machine server because it provides better response compared to the rest and can make use of the shared Web data.

Yes, there are circumstances in which a single-threaded server might be better. If it is designed such that:

- the server is completely CPU bound, such that multiple threads isn't needed. But it would account for some complexity that aren't needed.

An example is, the assistance number of a telephone directory (e.g 7771414) for an community of say, one million people. Consider that each name and telephone number record is sixty-four characters, the whole database takes 64 MB, and can be easily stored in the server's memory in order to provide quick lookup.

NOTE:

Multiple threads lead to operation slow down and no support for Kernel threads.

4 0

3 years ago

A microwave transmitter has an output of 0.1W at 2 GHz. Assume that this transmitter is used in a microwave communication system

Len [333]

Answer:

gain = 353.3616

P_r = 1.742*10^-8 W

Explanation:

Given:

- The output Power P_o = 0.1 W

- The diameter of the antennas d = 1.2 m

- The frequency of signal f = 2 GHz

Find:

a. What is the gain of each antenna?

b. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the available signal power out of the receiving antenna.

Solution:

- The gain of the parabolic antenna is given by the following formula:

gain = 0.56 * 4 * pi^2 * r^2 / λ^2

Where, λ : The wavelength of signal

r: Radius of antenna = d / 2 = 1.2 / 2 = 0.6 m

- The wavelength can be determined by:

λ = c / f

λ = (3*10^8) / (2*10^9)

λ = 0.15 m

- Plug in the values in the gain formula:

gain = 0.56 * 4 * pi^2 * 0.6^2 / 0.15^2

gain = 353.3616

- The available signal power out from the receiving antenna is:

P_r = (gain^2 * λ^2 * W) / (16*pi^2 * 10^2 * 10^6)

P_r = (353.36^2 * 0.15^2 * 0.1) / (16*pi^2 * 10^2 * 10^6)

P_r = 1.742*10^-8 W

4 0

3 years ago