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enyata [817]
3 years ago
7

A second order measuring system has a known natural frequency of 2000 Hz and damping ratio of 0.8. Would it be satisfactory to u

se it to measure a signal expected to have frequency content at up to 500 Hz? Demonstrate your answer by finding the expected dynamic error and phase shift at 500 Hz.

Physics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

The  expected dynamic error is  0.019

The phase shift is -23.10°C

Explanation:

The explanation is shown on the first uploaded image

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Will give brainliest! how does an engineer use physical science?
pentagon [3]

Answer: gravity, circuits

Explanation:

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3 years ago
3. The velocity of sound is 332 m/s. Answer the following questions:
Angelina_Jolie [31]

Answer:

20 Hz, 20000 Hz

0.0166 m, 16.6 m

Explanation:

The minimum frequency that a human ear can hear is 20 Hz

The maximum frequency that a human ear can hear is 20000 Hz.

v = Velocity of sound = 332 m/s

Wavelength is given by

\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{332}{20}\\\Rightarrow \lambda=16.6\ \text{m}

The longest wavelength that can be heard by the human ear is 16.6 m

\lambda=\dfrac{332}{20000}\\\Rightarrow \lambda=0.0166\ \text{m}

The shortest wavelength that can be heard by the human ear is 0.0166 m.

3 0
2 years ago
Newton
vodomira [7]

Given parameters:

Mass on earth  = 50kg

Unknown:

Mass on planet Xenon = ?

Weight on planet Xenon = ?

Mass is the amount of matter contained in a particular substance.

Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.

         Weight  = mass x acceleration due to gravity

Planet Xenon has half the gravitational force of Earth.

 This translated gives \frac{9.8}{2}   = 4.9m/s²

Now, mass is always the same every where if the amount of matter in a substance does not change.

In this problem, mass = 50kg on planet xenon.

Weight =  mass x acceleration due to gravity  = 50 x 4.9  = 245N

The weight on Xenon is 245N and the mass is 50kg

4 0
3 years ago
What is an example of radiation? check all that apply
Katen [24]
There are no appropriate examples in the list you provided with your question.

Examples of radiation:

... sunshine to tan your skin
... radio energy to bring you the news
... X-ray to check your teeth
... microwave to heat up the meatloaf
... flashlight to see where you're going
... RF energy to get an MRI of your knee
... infrared radiation from the campfire to warm your tootsies
... UHF radio waves to make a call or check Facebook with your smartphone
4 0
3 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
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