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3 years ago

What does frequency describe?

Physics

2 answers:

ddd [48]3 years ago

7 0

The whole definition of frequency is: <em>How often something happens. </em>

Especially referring to something that happens over and over and over and over.

One example is Choice-C: How often the particles of a medium vibrate.

"Frequency" comes from the word "frequent". That means "often", and "frequency" just means "often-ness" ... HOW often the thing happens.

Some other examples:

Frequency of jump-roping . . . maybe 60 per minute .

Frequency of rain . . . maybe 5 per month .

Frequency of an AM radio station . . . maybe 1 million waves per second.

(If it's something <u><em>per second</em></u>, then we call it "Hertz". That's not for the car rental company. It's for Heinrich Hertz, the German Physicist who was the first one to prove that electromagnetic waves exist. He sent radio waves all the way ACROSS HIS LABORATORY and detected them at the other side ( ! ), in 1887.)

Frequency of the wiggles in the sound wave coming out of a trumpet playing the note ' A ' . . . 440 Hertz.

Frequency of sunrise and the Chicago Tribune newspaper . . . 1 per day

Frequency of the cycle of Moon phases and an average human woman's ovulation cycle: 1 per 29.531 days, 1 per ~28 days .

Anarel [89]3 years ago

4 0

(C). How often the particles of a medium vibrate.

<h3>Further explanation</h3>

Let us analyze the answer choices one by one.

A: Distance from crest to crest or from trough to trough on a wave.

This statement represents the definition of wavelength. The distance between crest or between trough is the distance of one wave, that is the wavelength.

B: Depth of a wave.

The depth of a wave is the depth of influence of a water wave, which is about half the wavelength.

C: How often the particles of a medium vibrate.

Yes right, this statement describes frequency. The frequency of a wave refers to how often the particles of the medium vibrate when a wave passes through the medium.

D: Height of wave is the distance between two amplitudes.

Recall that the period and the frequency of any wave are inversely related.

$\boxed{ \ T = \frac{1}{f} \ } \rightarrow \boxed{ \ f = \frac{1}{T} \ }$

_ _ _ _ _ _ _ _ _ _

Definition:

Amplitude (A) = the maximum displacement from the mean position, in unit of meters (The SI measurement).
Period (T) = the time taken for one complete wave (The SI measurement: in second).
Frequency (f) = the number of oscillations completed per unit time, or the number of cycles per second (The SI measurement: in Hertz or Hz).
Wavelength (λ) = the shortest distance (in meters) along with the wave between two points that are moving exactly in step with one another. In other words, the wavelength is the distance traveled in one wave.
Wave speed (v) = the speed (in m/s) at which the wavefronts pass a stationary observer. The speed of the wave is the quotient of wavelength with the period or the product of wavelength with frequency.

$\boxed{ \ v = \frac{\lambda}{T} \ } \ \boxed{ \ v = \lambda f \ }$

<h3>Learn more</h3>

How are period and frequency related to each other? brainly.com/question/10239637
The characteristics of electromagnetic waves brainly.com/question/727976
Determine the shortest wavelength in electron transition brainly.com/question/4986277

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Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

Zarrin [17]

Answer:

a) (dT/dt) = -0.3 [T - 70]

b) (dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) (dT/dt) = -18 {T - [66 cos (2πt)]}

with t in hours

d) (dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Explanation:

The Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.

If the temperature of the object = T

Temperature of the surroundings = Ambient temperature = TA(t)

(dT/dt) ∝[T - TA(t)]

Introducing the constant of proportionality, k

(dT/dt) = k [T - TA(t)]

Temperature is in degree Celsius and time is in minutes.

Because the temperature of the body is decreasing, we introduce a minus sign

(dT/dt) = -k [T - TA(t)]

a) If TA(t) = 70°C, k = 0.3

(dT/dt) = -0.3 [T - 70]

b) The ambient temp TA(t) = 66 cos ((π/30)t) degrees Celsius (time measured in minutes).

(dT/dt) = -k [T - TA(t)]

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

(dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) If we measure time in hours the differential equation in part (b) changes.

1 hour = 60 mins

If t is now expressed in hours,

t hours = (60t) mins

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

dT = -k {T - [66 cos ((π/30)t)]} dt

dT = -k {T - [66 cos ((π/30)60t)]} d(60t)

(dT) = -60k {T - [66 cos ((π/30)60t)]} dt

(dT/dt) = -60k {T - [66 cos (2πt)]}

with t in hours, k = 0.3, 60k = 18

(dT/dt) = -18 {T - [66 cos (2πt)]}

d) If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more.

If T is in degree Fahrenheit

T°F = (5/9)(T°F - 32) degrees Celsius

T°F = [(5T/9) - 17.78] degrees Celsius

(dT/dt) = -60k {T - [66 cos (2πt)]}

time already converted to hours.

dT = -60k {T - [66 cos (2πt)]} dt

66 cos (2πt) degrees Celsius = {(9/5) [66 cos (2πt)] + 32} degrees Fahrenheit = {[118.8 cos (2πt)] + 57.6} degrees Fahrenheit

d[(5T/9) - 17.78] = -60k {T - [118.8 cos (2πt) + 57.6]} dt

(5/9) dT = -60k [T - 57.6 - 118.8 cos (2πt)] dt

(5/9) (dT/dt) = -60k [T - 57.6 - 118.8 cos (2πt)]

(dT/dt) = -108k [T - 57.6 - 118.8 cos (2πt)]

k = 0.3, 108k = 32.4

(dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Hope this Helps!!!

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4 years ago

A race car accelerates uniformly from 21.3 m/s to 26.5 m/s in 6.64 seconds. Determine the acceleration of the car and the distan

Sophie [7]

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3 years ago

3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.

Novay_Z [31]

Answer:

A. The bullet with 0.006kg has more energy

B. When the mass is doubled the kinetic energy increases

Explanation:

Kinetic energy increases when mass increases

kinetic energy increases when velocity increases

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3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$