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Anna35 [415]
3 years ago
8

Explain how each of the following terms relate to the Bernoulli equation: a. static pressure b. dynamic pressure .stagnation pre

ssure d. total pressure
Engineering
1 answer:
just olya [345]3 years ago
6 0

Explanation:

We know that  Bernoulli equation is the energy conservation equation.This equation given as

\dfrac{P}{\rho g}+ \dfrac{V^2}{2g}+Z=C

Where

P = Pressure

V= Velocity

Z= Elevation from reference

g= Acceleration due to gravity

ρ=Density

The above can be also written as

P+ \dfrac{\rho V^2}{2}+\rho gZ=C

ρ g Z = Static pressure

\dfrac{\rho V^2}{2}=Dynamic\ pressure

Stagnation pressure = Static pressure + Dynamic pressure

Stagnation\ pressure=\dfrac{\rho V^2}{2}+\rho gZ

The\ total\ pressure = \dfrac{P}{\rho g}+ \dfrac{V^2}{2g}+Z

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Answer:

35

Explanation:

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it's basically finding the perimeter of the square

so 35 is each side of the square that I am imagining in my head

so if you add 35...4 times you will get 140

for example:35

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3 years ago
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9.For a single-frequency sine wave modulating signal of 3 kHz with a carrier frequency of 36 MHz, what is the spacing between si
nikdorinn [45]

The spacing between sidebands is equal to 6 kHz.

<u>Given the following data:</u>

  • Modulating signal = 3 kHz.
  • Carrier frequency = 36 MHz.

<h3>What is a sideband?</h3>

A sideband can be defined as a band of frequencies that are lower or higher than the carrier frequency due to the modulation process. Thus, it will either be lower than or higher than the carrier frequency.

Generally, the frequency of the modulating signal is equal to the spacing between the sidebands. Therefore, a modulating signal of 3 kHz simply means that the lower sideband is <u>3 kHz</u> higher while the upper sideband is <u>3 kHz</u> lower.

Spacing = 3 kHz + 3 kHz = 6 kHz.

Read more on frequency here: brainly.com/question/3841958

8 0
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What are the potential hazards relating to materials handling injuries?
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3 years ago
Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e
marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

\Delta x=\frac{PL^3}{3EI}

Hence we can write

P=\frac{3EI\cdot \Delta x}{L^3}

Comparing with the standard spring equation F=kx we find the cantilever analogous to spring with k=\frac{3EI}{L^3}

Now the angular frequency of a spring is given by

\omega =\sqrt{\frac{k}{m}}

where

'm' is the mass of the load

Thus applying values we get

\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}

\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec

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3 years ago
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