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tiny-mole [99]
1 year ago
7

You are singing a song at karaoke and reach a part that requires a louder, more intense sound. What must you do to produce a lou

der sound
Physics
1 answer:
FrozenT [24]1 year ago
6 0

More energy is used in creating a louder sound from the mouth.

<h3>What must you do to produce a louder sound?</h3>

We use more energy in order to produce a louder sound because energy is the thing that helps in the formation of loud sound. Loudness is dependent on the energy that creates loud sound in the mouth. More energy we apply, more loud sound will produce. Energy is the main factor which gives us a loud sound in each and every instruments. Without, we can't imagine loud sound.

So we can conclude that more energy is used in creating a louder sound from the mouth.

Learn more about sound here: brainly.com/question/1199084

#SPJ1

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A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r
cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0
3 years ago
A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a
Slav-nsk [51]

For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.

For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

6 0
3 years ago
The wavelengths of radio waves are much _______________ than the wavelength of microwaves. therefore, radio waves carry much ___
nydimaria [60]
The wavelengths of radio waves are much "Longer" than the wavelength of microwaves therefore, radio waves carry much "Lower" <span>energy than a microwave.

Hope this helps!</span>
7 0
3 years ago
Read 2 more answers
When light moves from air to glass, part of the light is reflected and part of it is refracted. In the image, which ray shows re
marissa [1.9K]
You have not provided the diagram, therefore, I cannot provide an exact answer.
However, I will try to help by explaining how to solve this problem.

When light moves from air to glass:
1- part of the light is reflected back into the air where the angle of incidence is equal to the angle of reflection
2- part of the light enters the water and refracts. The angle of refraction can be calculated using Snell's law.

In a diagram, the reflected ray would be the one getting back into air while the refracted ray would be the one entering the water.

You can check the attached diagram for further illustrations.

Hope this helps :)

6 0
3 years ago
Read 2 more answers
Capacitors in Combination: A 5.0-μF, a 14-μF, and a 21-μF capacitor are connected in series. How much capacitance would a single
tankabanditka [31]

The total capacitance is <em>C</em> such that

1/<em>C</em> = 1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)

Solve for <em>C</em> :

<em>C</em> = 1 / (1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)) ≈ 3.1 µF

7 0
3 years ago
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