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3 years ago

What does resonance result in? Quieter sounds, frequency, amplitude

Physics

2 answers:

Amanda [17]3 years ago

6 0

Answer:

The answer is louder sounds. Resonance is the process, in which a system vibrates with increasing amplitudes, at specific frequencies of excitation. Resonance results in louder sounds. ... This is because of the surrounding environment, resonates with the cavity of the shell .

Explanation:

Tpy6a [65]3 years ago

4 0

Question:

What does resonance result in?

A) Quieter sounds B) frequency, C) amplitude D) Louder sounds.

Answer:

Resonance results in louder sounds.

Explanation:

In fact, the point at which the sound show maximum amplitude is termed as resonance. Since there is an increase in amplitude, the sound turns louder, and louder. Resonance is the situation, where one body vibrates after being effected by another vibrating body. Let us take an example of two bodies ‘A’ and ‘B’ at vibration.

Let the body A is in vibration at a particular frequency. And then, let the body ‘B’ disturbs with its vibration. Now, ‘A’ imparts some additional frequency, from ‘B’, which leads to resonance. So, if the amplitude increases, the sound produced is huge, and hence louder.

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Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t

kompoz [17]

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel $40\times 2=80 miles$

thus its position vector after two hours is

$r_A=-80sin35\hat{i}+80cos35\hat{j}$

similarly B travels with 20 mph and in 2 hours

$=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}$

Thus distance between A and B is

$r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}$

$|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}$

$|r_{AB}|=103.45 miles$

Velocity of A

$v_A=-40sin35\hat{i}+40cos35\hat{j}$

Velocity of B

$v_B=20sin80\hat{i}+20cos80\hat{j}$

Velocity of A w.r.t B

$v_{AB}=v_A-v_B$

$v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}$

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3 years ago

Which statements describe acceleration? Check all that apply. Negative acceleration occurs when an object slows down in the posi

garik1379 [7]

Answer:

Negative acceleration occurs when the acceleration vector points to the left.

1. Object slowing down in the positive direction.

2. Object speeding up in the negative direction.

Following six statements:

1. T

2. F

3. T

4. T

5. F

6. T

Check direction of acceleration vector.

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3 years ago

a surface recieving sound is moved from it original position to a position three times farther away from the source of the sound

Delicious77 [7]

Sound intensity = 1/(r^2)

That is Sound intensity is indirectly proportional to the distance. Therefore, sound becomes 9 times less intense.

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3 years ago

Why are peer reviews important?

riadik2000 [5.3K]

Peer review is important because it is used by scientists to decided which results should be published in a scientific journal

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3 years ago

Read 2 more answers

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

$\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r$

where,

$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

$\tau _{max}=\frac{4}{3}\times \frac{V}{A}$

Applying values we get

$\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa$

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3 years ago