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Mila [183]
3 years ago
6

The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?

Physics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e           I = \frac{Q}{t} ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

            Q = I × t

                = 1.5 × 2

               = 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

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

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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

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