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AlekseyPX
2 years ago
13

“I know what the atomic number of this atom is, but I don’t know what the number of electrons is,” a friend says. How would you

respond?
Physics
1 answer:
liq [111]2 years ago
8 0

Once the atomic number of an atom is known, the number of electrons can be deduced depending on if the atom is an ion or a neutral one.

<h3>Atomic number</h3>

The atomic number of an atom is the number of protons in the nucleus of the atom.

For atoms that are neutral, that is, no net charges, the number of protons is always equal to the number of electrons. In other words, the positive charges always balance the negative charges in neutral atoms.

Thus, if the atomic number of a neutral atom is 6, for example, the proton number will also be 6. Since the proton must balance the electron, the number of electrons will also be 6.

More on atomic numbers can be found here; brainly.com/question/17274608

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A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
The speed of light in a material is found to be 2.07 x 108 m/s. What is the most likely material from the options below if the s
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Answer:

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Explanation:

Write the expression for the refractive index.

n=c/v

c= speed of light in air

v= speed of light in material

=(3×10^8 m/s)/(2.2×10^8 m/s)

 =1.3636

4 0
4 years ago
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Answer:

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Explanation:

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