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EastWind [94]
3 years ago
12

I need the right answer ASAP NO LINKS!!!

Physics
1 answer:
serious [3.7K]3 years ago
3 0

Answer:

Water and power come from external sources.

Explanation:

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A car manufacturer claims an enginerring breakthrough, since it successfully reduced its interior noise level from 65dB to 55 dB
Stells [14]

Yes, that is truly amazing.

They significantly reduced the noise's intensity by 90%, that might be proper reaction.

<h3>What is noise?</h3>
  • Unwanted sound that is regarded loud, unpleasant, or disruptive to hearing is called noise.
  • Physically speaking, there is no difference between unwanted sound and desired sound because both are vibrations traveling through a medium like air or water.
  • When the brain receives and interprets a sound, a difference occurs.
  • Decibels are used to measure sound (dB).
  • A motorcycle engine operating is roughly 95 dB louder than regular conversation, which is around 60 dB louder than a whisper.
  • Your hearing may begin to be harmed if exposed to noise over 70 dB for an extended period of time.
  • Your ears can suffer instant damage from loud noise above 120 dB.
<h3>What is sound intensity?</h3>
  • The power carried by sound waves per unit area in a direction perpendicular to that region is known as sound intensity or acoustic intensity.
  • The watt per square meter is the SI measure of intensity, which also covers sound intensity.

Learn more about sound here:

brainly.com/question/14595927

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5 0
2 years ago
Critical Thinking 1. One day, a group of neighbourhood degs Started barking on their own. The residents living in that area are
liberstina [14]

The reason behind the barking of the group of neighborhood dogs could be <u>Territorial/Protective</u> or  <u>Alarm/Fear</u>.

What is bark or barking?

Dogs are known for making the most prevalent barking noises. Wolves, coyotes, foxes, seals, and barking owls are a few other creatures that make this sounds. The most typical onomatopoeia for this sound in English, particularly for large dogs, is "woof." The word "bark" can also be used to describe the sound made by numerous canids. A bark is a brief vocalization, as per the researchers at the University of Massachusetts Amherst.

Territorial/Protective: When a person or another animal enters their perceived area, your dog may bark excessively, depending on the breed. The barking frequently gets louder as the threat draws closer. When your dog barks in this manner, he will appear vigilant and sometimes even hostile.

Alarm/Fear: Some dogs will bark in alarm or fear at any sound or item that draws their attention or alarmed them. It's not just in their home country where this can occur. When they are afraid, their tails will be tucked and their ears will be pulled back.

To know more about bark, go to link

brainly.com/question/5082728

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3 0
1 year ago
An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.
UNO [17]

Explanation:

The given data is as follows.

        Electric field strength (E) = 187,500 V/m

    Magnetic field strength (B) = 0.125 T

       Diameter (d) = 25.05 cm = 0.2505 m    (as 1 m = 100 cm)

    Radius (r) = \frac{d}{2}

                    = \frac{0.2505}{2}

                    = 0.12525 m

Formula to calculate the magnetic force (F_{M}) is as follows.

              F_{M} = Bqv ............ (1)

Electrical force is calculated as follows.

             F_{E} = qE ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

       F_{M} - F_{E} = 0

or,             F_{M} = F_{E}

Hence, from equations (1) and (2)

              Bqv = qE

or,            v = \frac{E}{B} ............. (3)

                  = \frac{187500 V/m}{0.125 T}

                  = 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

              F_{c} = \frac{mv^{2}}{r} ........... (4)

where,    F_{c} = centripetal force

             F_{M} = F_{c}

Using equation (1) and (4) as follows.

            F_{M} = F_{c}

              Bqv = \frac{mv^{2}}{r}

                   \frac{q}{m} = \frac{v}{Br}

                       = \frac{15 \times 10^{5}}{0.125 \times 0.12525}

                       = 958.08 \times 10^{5} C/kg

Thus, we can conclude that charge-to-mass ratio of the given particle is 958.08 \times 10^{5} C/kg.

8 0
3 years ago
The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th
ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, B=2\ mT=2.2\times 10^{-3}\ T

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :                  

B=\dfrac{\mu_oNI}{2\pi r}

r=\dfrac{\mu_oNI}{2\pi B}  

r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}  

r = 0.00523 m

or

r=5.23\times 10^{-3}\ m

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
What causes the phases of our moon
Helga [31]

sunlight, and the position of the earth

4 0
4 years ago
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