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3 years ago

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Dawn and Timothy are sailboating in Lake Obewon. Starting from rest near the shore, they accelerate with a uniform acceleration

of 0.39 m/s/s. How far are they from the shore after 18 seconds? *

1 answer:

castortr0y [4]3 years ago

5 0

Answer:

The value is $s =63.18 \ m$

Explanation:

From the question we are told that

The uniform acceleration is $a = 0.39 \ m/s^2$

The time considered is $t = 18 \ seconds$

Generally the from kinematic equation we have that

$s = ut + \frac{1}{2} at^2$

Given that the boat started from rest , the initial velocity is u = 0 m/s

So

$s = 0 * 18 + \frac{1}{2} *0.39 * 18^2$

=> $s =63.18 \ m$

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An object is thrown directly up (positive direction) with a velocity (v o ) of 20.0 m/s and d o = 0. How high does it rise (v =

Lelechka [254]

Answer:

h=20.38 m

Explanation:

Given that

Initial speed of object u = 20 m/s

Acceleration $g= 9.8\ \frac{m}{s^2}$

We know that

$v^2=u^2-2gh$

Here acceleration and velocity is in opposite direction so the object will come rest after reaching at distance h.When body will reach at its highest position then velocity will become zero(v=0).

Now by putting the values

$v^2=u^2-2gh$

$0 =20^2-2\times 9.1\times h$

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3 years ago

If an oxygen molecule traveling at the rms speed bounces back and forth between opposite sides of a cubical vessel of 0.10 m on

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3 years ago

Is energy made or lost over time

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4 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

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3 years ago

Two identical loudspeakers that are 5.00 m apart and face toward each other are driven in phase by the same oscillator at a freq

Lunna [17]

Answer:0.0982 m

Explanation:

Given

distance between two loudspeaker is 5 m

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speed of sound (v)=344 m/s

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$x_1=\frac{L}{2}-x_0$

Position of observer w.r.t to second speaker is

$x_2=\frac{L}{2}+x_0$

$\Delta x=2x_0$

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$\Delta x=\left ( m+\frac{1}{2}\right )\cdot lambda$

For minimum m=0

and $\lambda =\frac{v}{f}$

$2x_0=\frac{v}{2f}$

$x_0=\frac{v}{4f}=\frac{344}{4\times 875}=0.0983 m$

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3 years ago