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2 years ago

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If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these

engines?
a) 44 dB
b) 67 dB
c) 13 dB
d) 88 dB

1 answer:

insens350 [35]2 years ago

5 0

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

$\beta = 10Log(\frac{I}{I_0})$

Where,

I = Intensity of Sound

$I_0$ = Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

$I= 15I_0$

And the total mutual intensity is 100 dB, so we should

$\beta = 100-10*log(\frac{15I_0}{I_0})$

$\beta = 100-10*log(15)$

$\beta = 100-11.76$

$\beta = 88.23dB$

Therefore each one of these engines produce D. 88dB.

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Answer:

14778.29 N

Explanation:

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Since stress, $\sigma= \frac {F}{A}$ where F is force, A is area and since specimen is cylindrical,

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Also strain $\epsilon= \frac { \triangle L}{L}$ where L is length

Poison’s ratio,v is the ratio of lateral strain to longitudinal strain hence

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2 years ago

The density of aluminum is 2.7 × 103 kg/m3 . the speed of longitudinal waves in an aluminum rod is measured to be 5.1 × 103 m/s.

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- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

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We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

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So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

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(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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2 years ago

While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6

Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

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2 years ago