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NARA [144]
3 years ago
14

The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

Physics
2 answers:
zlopas [31]3 years ago
8 0

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

  • Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency
marishachu [46]3 years ago
8 0

Answer:

The change in intensity is 63%.

Explanation:

intensity level = 18 db

Let the intensity is I.

Io = 10^(-12) W/m^2

Use the formula of intensity

dB = 10 log\left ( \frac{I}{Io} \right )\\\\18 = 10 log\left ( \frac{I}{Io} \right )\\\\1.8 = log\left ( \frac{I}{Io} \right )\\\\\left ( \frac{I}{Io} \right )=63.1

So, the change in intensity is 63%.

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Fraternity hazing is acceptable because it is an initational rite to the brotherhood
Fantom [35]

Answer:

that is the right answer

Explanation:

8 0
3 years ago
Volume equals 4÷3 times pi times nine to the power of 3
Tcecarenko [31]

Answer:

V=972π

The equation I used... V=4/3π(9)^3

4 0
2 years ago
Find the sum of the vectors: 40m/s2 Northeast, 10 m/s2 Northeast
podryga [215]

Answer:

Explanation:

Since both vectors are pointing on the same direction (Northeast), the sum of them will point in that same direction, and its magnitud will be the sum of the magnitudes of each vector (40m/s2+10m/s2). This problem is just a problem in one dimension. The sum of the vectors is then 50m/s2 Northeast.

8 0
4 years ago
What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incide
Softa [21]

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

<u>c. The incident light must have at least as much energy as the electron work function</u>

3 0
3 years ago
What is the potential energy of a 50 newton object resting on the edge of a cliff 30 meters high?
mojhsa [17]
The potential energy of the object is going to be gravitational: PE=mgh. Assuming we're talking about the gravitational potential energy relative to the bottom of the cliff, the object's height is 30 m. g is 9.8 m/s^2. We don't know its mass. You could technically use the equation F=mg to find the mass, where F=40 and g=9.8, but that's unnecessary. You can just substitute F into PE=mgh to get PE=Fh. Substitute the given values to get PE=40*30=120 J.
3 0
3 years ago
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