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4 years ago

What is resonance frequency?

2 answers:

7 0

Resonance frequency; Online Definition:
In physics resonance describes the phenomenon of amplification that occurred when the frequency of a periodically applied force is harmonic proportion to a natural frequency of a system
[ phenomenon in which a vibrating system or external force drives another system to oscillate with greater amplitudes at specific frequencies ]

My definition:
Natural frequencies of vibrations determined by meters from another vibrating object :)

Bogdan [553]4 years ago

5 0

A natural frequency of vibrations determined by the physical parameters of the vibrating object.

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Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an al

olga nikolaevna [1]

Answer:

$6.67154\times 10^{-9}\ F$

13.009503 C

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

k = Dielectric constant of air $3\times 10^6\ V/m$

Side of plate = 0.7 km

A = Area

d = Distance = 650 m

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F$

The capacitance is $6.67154\times 10^{-9}\ F$

Electric field is given by

$Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C$

The charge on the cloud is 13.009503 C

7 0

3 years ago

A room that has an average ambient sound pressure level of 62 dBA and a maximum sound pressure level lasting more than a minute

cupoosta [38]

Answer:

Explanation:

A fire alarm notification appliance is an active fire protection component of a fire alarm system. The primary function of the notification appliance is to alert persons at risk.

If want the audible public mode signal to be hear clearly then, we need to have a sound level that is at least 15dB above the average ambient sound level or 5dB above the maximum sound level of at least 1minute

In this case the,

The average ambient sound level is 62dB,

And the maximum sound level is 68dB

Then, the public mode signal should be at least

1. 62dB+ 15dB=77dB

2. 68dB +5dB =73dB.

Then the public mode signal hearing must be at least 77dB.

6 0

4 years ago

Which type of electromagnetic radiation is most likely to be ionizing

zlopas [31]

Answer:

Xrays. i hope that helps

Explanation:

4 0

3 years ago

Read 2 more answers

Before leaving her home, Myesha makes sure that every electrical appliance is unplugged and checks that every window and door is

Kryger [21]

Answer:Obsessive-Compulsive Disorder

Explanation:

Obsessive-Compulsive Disorder is a general, persistent, and soon-lasting disorder in which an individual has involuntary, repetitive thoughts and/or actions that he or she feels compelled to replicate over and over.

Myesha also shows the symptoms of OCD by constantly checking the window and doors.

An individual suffering from OCD shows symptoms like

Fear of or infection by germs
Aggressive feelings to others or self
Getting things in a symmetrical or optimal order

8 0

3 years ago

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi

zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} = mass flow rate\\\rho = Density\\V = Velocity$

values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3V_1 = 40m/s$

$A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s$

PART A) Applying the flow equation

$\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2$

7 0

3 years ago