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serious [3.7K]
3 years ago
6

calculate the temperature at which the reading on Fahrenheit scale is equal to half the reading of Celsius scale​

Physics
1 answer:
RideAnS [48]3 years ago
3 0

Answer:

Therefore, the temperature at which the Fahrenheit scale reading is equal to half of the Celsius scale is −24.6∘C .

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entropy

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
Consider the line graph depicting the motion of two cars. What statement regarding the cars motion is accurate?
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Answer:

Its C

Explanation:

I did it on usatestprep

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3 years ago
Sequence eye light enters the eye
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Well it is the courtnes

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3 years ago
A 140 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless trac
photoshop1234 [79]

Answer:

Explanation:

mass attached m = .14 kg

force constant k = 5N / m

displacement

= amplitude of oscillation

A = .03 m

A ) period of motion = 2\pi\sqrt{\frac{m}{k} }

= 2 x 3.14 \sqrt{\frac{.14}{5} }

T = 1.05 s

B ) maximum speed of block = angular velocity x amplitude

= (2π /T)  x A

= (2 x 3.14 x .03) / 1.05

= .1794 m / s

17.94 cm /s

C )

maximum acceleration = angular velocity² x amplitude

= (2π /T)² x A

= (2π /1.05)² x .03

= 1.073 m / s²

D )

position

S = A cos ωt , ω is angular velocity

S = .03 cos(2πt /T)

= .03 cos 5.98 t

v =ω A sin(2πt /T)

= 5.98 x .03 sin5.98t

= .1794 sin5.98t

acceleration = ω²A sin5.98t

= 1.073 sin5.98t

7 0
3 years ago
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