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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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Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo
kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

W=F*d*cos(\theta)

here we have the work done by Paul and the friction force, so:

W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J

W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J

Now the change of energy is:

W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s

4 0
3 years ago
Read 2 more answers
What is the current if 4C of charge passes in 2 s?
julia-pushkina [17]

Answer:

I hope 2 amperes of current passes

8 0
2 years ago
2.
horrorfan [7]

Answer:

B. 6HgO → 6Hg + 3O_{2}

Explanation:

A decomposition reaction is a reaction in which a single reactant is broken down into 2 or more products.

6 0
3 years ago
Ana walks from 4 m to 200 cm. Which of the following statements is true about
Gemiola [76]
Distance=2m

because 200cm = 2m
so 4m-2m=2m
3 0
3 years ago
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I
Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = \frac{4\times \pi^2\times r^3}{GM} where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

8 0
3 years ago
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