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3 years ago

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A transverse wave is moving toward the right in a uniform medium. Point X represents a particle of the uniform medium. Which dia

gram represents the direction of the motion of particle X at the instant shown?

2 answers:

trapecia [35]3 years ago

6 0

After looking at the transverse waves in the diagram you listed above, the one diagram that does represent the direction of particle X at the instant show in diagram number 3. The direction of the wave motion is up. The correct answer choice will be 3.

vladimir2022 [97]3 years ago

6 0

<h3><u>Answer</u>;</h3>

(4) last diagram

<h3><u>Explanation</u>;</h3>

<em><u>A transverse wave is a type of wave where the movement of the particles of the medium is perpendicular to the direction of the propagation of the wave. </u></em>
Transverse waves forms regions of maximum displacement known as crests and troughs. The crests are the top points of the wave and the trough is the bottom point of the wave.
In a transverse wave particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.

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A gas had a initial volume of 168 cm at a temperature of 255 k and a pressure of 1.6 atm the pressure of the gas decreased to 1.

Mademuasel [1]

P₁V₁ / T₁ = P₂V₂ / T₂

1.6 × 168 /255 = 1.3 × V₂ / 285

V₂ = 1.6 × 168 × 285 / (1.3 × 255)

V₂ = 231.095

The final volume = 231 cm³

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3 years ago

What is meant by lateral inversion

grandymaker [24]

Answer:

Explanation:

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3 years ago

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the d

frez [133]

The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

$\beta=10log(\frac{I}{I_{0}} )dB$ ,

where $I_{0}$ is a reference intensity. for sound waves, $I_{0}$ is taken to be $10^{-12} W/m^{2}$ . Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. $I=10I_{0}$ ? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. $I=100I_{0}$ ? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels ( $\Delta \beta_{2},\Delta \beta_{4}$ and $\Delta \beta_{8}$ ) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) $I=10I_{0}$

$\beta=10log(\frac{10I_{0}}{I_{0}} )$

$\beta=10log(10 )$

β = 10

<u>The sound Intensity level with intensity 10x is </u><u>10dB</u>.

B) $I=100I_{0}$

$\beta=10log(\frac{100I_{0}}{I_{0}} )$

$\beta=10log(100)$

β = 20

<u>With intensity 100x, level is </u><u>20dB</u>.

C) To calculate the change, take the f to be the factor of increase:

For $\Delta \beta_{2}$ :

$I=2I_{0}$

$\beta=10log(\frac{2I_{0}}{I_{0}} )$

$\beta=10log(2)$

β = 3

For $\Delta \beta_{4}$ :

$I=4I_{0}$

$\beta=10log(\frac{4I_{0}}{I_{0}} )$

$\beta=10log(4)$

β = 6

For $\Delta \beta_{8}$ :

$I=8I_{0}$

$\beta=10log(\frac{8I_{0}}{I_{0}} )$

β = 9

Change is

$\Delta \beta_{2},\Delta \beta_{4}$ , $\Delta \beta_{8}$ = 3,6,9 dB

6 0

3 years ago

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kirza4 [7]

D. There's not enough research surrounding it. A big question is if they have side effects due to genetic modification that we don't know of.

(Not sure if you had D eliminated or? But it should be the correct answer)

5 0

3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago