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2 years ago

The period of an 800 hertz sine wave is

Engineering

1 answer:

sukhopar [10]2 years ago

3 0

Explanation:

White Board Activity: Practice: A sound has a frequency of 800 Hz. What is the period of the wave? The wave repeats 800 times in 1 second and the period of the function is 1/800 or 0.00125.

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Give me the description of - Feedforward control loops with an example.

ValentinkaMS [17]

Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

Explanation:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

example of feedforward is

Shower

which consist of following control points

Hear toilet flush (measurement)

Customize water to compensate

feedback refers to that point when water turns hot before the configuration changes

5 0

3 years ago

water flows in a horizontal constant-area pipe; the pipe diameter is 75 mm and the average flow speed is 5 m/s. At the pipe inle

Veronika [31]

Answer:

Head loss = 28.03 m

Explanation:

According to Bernoulli's theorem for fluids we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=Constant$

Applying this between the 2 given points we have

$\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{l}$

Here $h_{l}$ is the head loss that occurs

$\therefore h_{l}=\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}-\frac{P_{2}}{\gamma _{w}}-\frac{V_{2}^{2}}{2g}-z_{2}$

Since the pipe is horizantal we have $z_{1}-z_{2}=0$

Applying contunity equation between the 2 sections we get

$A_{1}V_{1}=A_{2}V_{2}\\\\\therefore V_{1}=V_{2}(\because A_{1}=A_{2})$

Since the cross sectional area of the both the sections is same thus the speed

is also same

Using this information in the above equation of head loss we obtain

$h_{l}=\frac{1}{\gamma _{w}}(P_{1}-P_{2})$

Applying values we get

$h_{l}=\frac{1}{9810}\times (275\times 10^{3})m\\\\\therefore h_{l}=28.03m$

3 0

3 years ago

The gas velocity decrease from 480 m/s to 160 m/s through a stationary normal shock. If the pressure and density upstream of the

Ainat [17]

Answer:

52m/s

12 kpa

0.1 kg/m

Explanation:

make me brainless tyyy

6 0

2 years ago

Determine the resistance of 3km of copper having a diameter of 0,65mm if the resistivity of copper is 1,7x10^8

LUCKY_DIMON [66]

Answer:

Resistance of copper = 1.54 * 10^18 Ohms

Explanation:

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

Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m

Resistivity, P = 1.7 * 10^8 Ωm

Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m

$Radius, r = \frac {diameter}{2}$

$Radius = \frac {0.00065}{2}$

Radius = 0.000325 m

To find the resistance;

Mathematically, resistance is given by the formula;

$Resistance = P \frac {L}{A}$

Where;

P is the resistivity of the material.
L is the length of the material.
A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of copper.

Area of circle = πr²

Substituting into the equation, we have;

Area = 3.142 * (0.000325)²

Area = 3.142 * 1.05625 × 10^-7

Area = 3.32 × 10^-7 m²

Now, to find the resistance of copper;

$Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}}$

$Resistance = 1.7 * 10^{8} * 903614.46$

<em>Resistance = 1.54 * 10^18 Ohms </em>

3 0

3 years ago

A wastewater treatment plant discharges 2.0 m^3/s of effluent having an ultimate BOD of 40.0 mg/L into a stream flowing at 15.0

dexar [7]

Answer:

What grade is this is for??

4 0

3 years ago