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3 years ago

10

Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is

the new level of sound β2?

1 answer:

topjm [15]3 years ago

4 0

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

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A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i

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Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

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Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)

$V=V_{o} + at$ (2)

Where:

$y$ is the <u>height of the model rocket at a given time</u>

$y_{o}=0$ is the i<u>nitial height </u>of the model rocket

$V_{o}=0$ is the<u> initial velocity</u> of the model rocket since it started from rest

$V$ is the <u>velocity of the rocket at a given height and time</u>

$t$ is the <u>time</u> it takes to the model rocket to reach a certain height

$a$ is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)

Finding $a$ :

$a=86.847 m/s^{2}$ (10)

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$ :

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)

$V= (86.847 m/s^{2})(0.2 s)$ (14)

Finally:

$V=17.369 m/s$ (15)

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