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Alenkinab [10]
3 years ago
6

Which best describes the way a sound wave is sent trough the radio?

Physics
2 answers:
Fittoniya [83]3 years ago
5 0

Answer:

D

Explanation:

horsena [70]3 years ago
4 0
Sound wave > electric signal > radio wave > sound wave
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In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
Mazyrski [523]

Answer:

54 g of oxygen ...

4 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

7 0
3 years ago
How can you relazie a perfect balck body in pratice​
Mazyrski [523]
A perfect black body can’t be realized
5 0
3 years ago
Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.
Gnoma [55]

Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

6 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
4 years ago
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