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sdas [7]
3 years ago
6

Please Help! If you send a sound wave of the same wavelength (_ = 2.00 m) through air, helium, and carbon dioxide, describe how

the pitch of the sound will compare through each medium. Use calculations and your data to explain
Physics
2 answers:
IrinaVladis [17]3 years ago
8 0

Answer: The pitch would be maximum in Helium then air and then in carbon dioxide.

Explanation:

The speed of wave, frequency and wavelength are related as:

f = v/λ

Since the wavelength is constant in all the mediums, the frequency is directly proportional to the speed of the wave in the medium.

Pitch is dependent on frequency, greater the frequency, higher is the pitch.

Speed of wave in air at room temperature = 343 m/s

The frequency of wave in air ⇒ f =  343 m/s/2.0 m = 171.5/s

Speed of wave in Helium at room temperature = 1007 m/s

The frequency of wave in Helium⇒ f =  1007 m/s/2.0 m = 503.5/s

Speed of wave in carbon-dioxide at room temperature = 267 m/s

The frequency of wave in carbon-dioxide ⇒ f =  267 m/s/2.0 m = 133.5 /s

Thus, since the speed of sound is greatest in Helium, it has greatest frequency and hence highest pitch as compared to air. Pitch would be least in carbon-dioxide.

Andreas93 [3]3 years ago
7 0
<span>The density of air is about 1.29 kg/m^3, of helium .169 kg/m^3, and of carbon dioxide 1.98 kg/m^3. So the pitch would be highest for helium, in the middle for air, and lowest for CO2.</span>
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Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth o
postnew [5]
The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
    = (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
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Answer: 539.7 m²

8 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
How is the medium of a wave and transmit of a wave similar and different?
zepelin [54]

Answer:

Sound and light are similar in that both are forms of energy that travel in waves. They both have properties of wavelength, freqency and amplitude. Here are some differences: Sound can only travel through a medium (substance) while light can travel through empty space.

3 0
3 years ago
An ideal air-filled parallel plate capacitor with plate a separation of 4.0 cm has a plate area of 0.040 m2. what is the capacit
irga5000 [103]

An ideal air-filled parallel plate capacitor with plate a separation of 4.0 cm has a plate area of 0.040 m2. what is the capacitance of this capacitor with air between these plates<u> 8.9 pF.</u>

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates.

The capacitance of a parallel plate capacitor depends on area of each plate, dielectric medium between the plates and distance between the plates.

The amount of energy stored in a plate capacitor is given by

⇒ U = \frac{Q^{2}}{2C},

where, Q is the stored charge and C is the capacitance,

To learn more about parallel plate capacitor, here

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2 years ago
In the figure, which arrow represents the amplitude of the pendulum's motion?
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That's arrow 'd' in the drawing.
3 0
4 years ago
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