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2 years ago

For which job would a noise canceling device be the most beneficial?

Physics

2 answers:

musickatia [10]2 years ago

8 0

A carpenter using saws and cutting tools as the tools make loud noises.

Alex Ar [27]2 years ago

6 0

Answer:

a carpenter using saws and cutting tools

Explanation:

In noise cancelling system we know that two sounds are cancelled just by using the concept of destructive interference.

Here when two sounds which are in opposite phase are superimposed with each other then due to this superposition of waves the resultant intensity will be less than the intensity of any individual wave.

This phenomenon is known as destructive interference.

This is used to decrease the level of sound in such a part of work where we have large sound in output so here correct answer will be

a carpenter using saws and cutting tools

This is because it produce large noise while we use this cutting tools

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Experiment: Group 1 drinks 500 mL of coffee a day.

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How does the drink content affect an individual's blood pressure?

Explanation:

In every experiment using the scientific method, an observation lays the foundation of that experiment. A problem must be observed, which then leads to asking a SCIENTIFIC QUESTION in order to investigate. A scientific question must include the variable being changed called INDEPENDENT VARIABLE and the variable being measured called DEPENDENT VARIABLE.

In this experimental procedure or set up,

- Group 1 drinks 500 mL of coffee a day.

- Group 2 drink 500 mL of tea a day,

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This set up indicates that the variable being changed (independent) is the DRINK CONTENT while the variable being measured (dependent) is the BLOOD PRESSURE. Hence, these variables serve as the template to ask a scientific question which goes thus:

HOW DOES THE DRINK CONTENT AFFECT AN INDIVIDUAL'S BLOOD PRESSURE?

This scientific question relates how the independent variable (drink) causes the dependent variable to respond (blood pressure).

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2 years ago

What is meant by the statement "Motion is relativel"?

aleksklad [387]

Answer:

when an object move from its place is called an mation

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3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

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2 years ago