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2 years ago

Explain how you can recognize the sound of a flute

Physics

2 answers:

Anton [14]2 years ago

7 0

I agree with the other person

lions [1.4K]2 years ago

5 0

Answer:

It's airy, light, poetic, mellow, bright, wafting, ethereal, rich, soft, graceful, penetrating, brilliant, clear, shrill, silvery, wind-like, whistling, whispering, humming, filigree, sighing, aspirate.

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What is someone who leaves there country to settle in another?

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Answer:

an immigrant

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al aplicar una fuerza de 2 N sobre un muelle este se alarga 4cm.¿cuanto se alargara si la fuerza es el triple?¿que fuerza tendri

3241004551 [841]

1) 12 cm

2) 3 N

Explanation:

The relationship between force and elongation in a spring is given by Hooke's law:

$F=kx$

where

F is the force applied

k is the spring constant

x is the elongation

For the spring in this problem, at the beginning we have:

$F=2 N$

$x=4 cm$

So the spring constant is

$k=\frac{F}{x}=\frac{2N}{4 cm}=0.5 N/cm$

Later, the force is tripled, so the new force is

$F'=3F=3(2)=6 N$

Therefore, the new elongation is

$x'=\frac{F'}{k}=\frac{6}{0.5}=12 cm$

In this second problem, we know that the elongation of the spring now is

$x=6 cm$

From part a), we know that the spring constant is

$k=0.5 N/cm$

Therefore, we can use the following equation to find the force:

$F=kx$

And substituting k and x, we find:

$F=(0.5)(6)=3 N$

So, the force to produce an elongation of 6 cm must be 3 N.

6 0

3 years ago

A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1

stellarik [79]

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = $\frac{4}{3}\pi r^3$

M = Molar mass of helium = $4.0026\times 10^{-3}\ kg/mol$

$\rho$ = Density of surrounding air = $1.19\ kg/m^3$

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

$mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg$

Mass of helium is 6.4356 kg

Moles of helium

$n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489$

Ideal gas law

$P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa$

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0

3 years ago