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2 years ago

Explain what happens to the pitch of a cell phone ring when the wavelength of a sound wave increases

Physics

2 answers:

max2010maxim [7]2 years ago

4 0

You get a more low sound.

Conversely, when the wavelength becomes shorter you get a more treble sound.

;-)

andrew-mc [135]2 years ago

4 0

"Pitch" of sound is what we hear as the result of its frequency.

When the wavelength of any sound INcreases, its frequency DEcreases.

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The teachers of Scott, Christopher, Dianne, and Kailee have last names Koeninger, Dannemiller, Briscoe, and Carter. Match the ki

Pani-rosa [81]

Scott-Dannemiller, Koeninger, Briscoe, and Carter

Christopher-Briscoe, Dannemiller, and Koeninger

Dianne-Koeninger, Briscoe, and Carter

Kailee-Koeninger and Carter

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3 years ago

Which type of line is shown on the map

emmainna [20.7K]

Answer:

Explanation:

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2 years ago

Two examples of an insulator

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Some examples of insulators are plastic, glass and rubber.

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3 years ago

What is the approximate energy required to raise the temperature of 1.00 L of hydrogen by 90 °C? The pressure is held constant a

Oliga [24]

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

$Q = n C_p \Delta T$

here we know that

n = number of moles

$n = \frac{1}{22.4}$

$n = 0.0446$

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

$C_p = \frac{7}{2}R$

now we have

$Q = (0.0446)(\frac{7}{2}R)(90)$

$Q = 116.8 J$

3 0

2 years ago

Ask Your Teacher Two long, straight wires are parallel and 11 cm apart. One carries a current of 2.9 A, the other a current of 5

dsp73

Answer:

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi\times 10^{-7}\ N/A^2$

$i_1$ = Current in first wire = 2.9 A

$i_2$ = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

7 0

2 years ago