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3 years ago

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In a football game, a defensive lineman exerts a force upon an offensive lineman, causing the offensive lineman to be accelerate

d backward at 2 m/s2. If the offensive lineman's mass were doubled, how much force would the defensive lineman have to apply to accelerate the offensive lineman 2 m/s2 backward? A. half as much B. twice as much C. four times as much D. the same amount

2 answers:

Grace [21]3 years ago

7 0

the answer is B.twice as much

skelet666 [1.2K]3 years ago

7 0

Answer:

The answer is B (twice as much)

Explanation:

The magnitude, or amount, of an object's acceleration is directly proportional to the magnitude of the force exerted upon it and inversely proportional to its own mass.

In this case, if the offensive lineman's mass were doubled, then the force exerted upon him would also have to be doubled to produce the same acceleration (2 m/s2).

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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

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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.

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

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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:

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

$\beta=10log(2)$

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For $\Delta \beta_{4}$ :

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

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For $\Delta \beta_{8}$ :

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

β = 9

Change is

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

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