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2 years ago

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if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate

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1 answer:

777dan777 [17]2 years ago

4 0

The sound power the person generated is $2.313 \times 10^{-5} \ W$ .

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

$A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2$

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

$P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W$

Learn more about intensity of sound here: brainly.com/question/4431819

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(1) find the density of a substance if the mass of the substance is 150kg and dimension 20mby10mby5m

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Answer:

0.15kg/m³

Explanation:

Density = mass/ volume

Given that

Mass = 150kg

Note that volume = length x breadth x height

Volume = 20 x 10 x 5

Volume = 1000m³

Density = mass ➗ volume

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Density = 0.15kg/m³

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3 years ago

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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