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Dmitry_Shevchenko [17]
3 years ago
8

In which of these cases do we have enough information to say that the atom is electrically neutral?

Physics
2 answers:
Luba_88 [7]3 years ago
5 0
<span>So we want to know when do we have enough information about the atom that we can say that it is electrically neutral. So the information we need to know is the number of protons, whic are positively charged particles and the number of electrons which are negatively charged particles. Number of protons has to be equal to number of electrons for the atom to have neutral charge. Neutrons are electrically neutral so their number doesn't change the charge of the atom. So the correct answer is B. </span>
Svetradugi [14.3K]3 years ago
3 0
B. 7 protons and 7 electrons, just took the test
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Its a waste of time, you have to not only write it down, but study it after too . other than that notes are great.

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Relay of thermostate is not working why
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7 0
2 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
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HELP I NEED THE ANSWER NOW
Setler79 [48]

Answer:

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Explanation:

The famous Einstein's equation is:

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In this equation, Einstein summarized the following fact: mass can be converted into energy, and the amount of energy released in such a process is given by the equation.

An example of application of this equation is the nuclear fusion process. In a nuclear fusion, two lighter nuclei combine into a heavier nucleus. However, the mass of the heavier nucleus is slightly less than the sum of the masses of the two original nuclei: some of the mass of the original nuclei has been converted into energy, accorging to the previous equation.

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cricket20 [7]

Answer:

Nuclease is the answer I know

I hope this is the answer

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