Answer: For number 9, the children would not be able to close the box.
Explanation: The combined force of the children is 55 N. The box spring overpowers them with 60 N. I hope this helps!
An electron shell can hold 2(n^2) electrons (technically) where n is the shell number, i.e. shell 1 can hold 2, shell 2 can hold 8, 3 holds 18 and so on.
The atomic number of Nitrogen is 7, i.e. it has 7 electrons (to match its 7 protons, assuming it isn't an ion).
With the atomic number, you simply start from shell 1 and work out. So we put 2 electrons in shell 1, leaving us with 5 left. Shell 2 can hold 6 so we can fit all 5 in.
In other words, you should have 2 electron shells on the atom, shell 1 with 2 e- and shell 2 with 5 e-.
The answer is E. Short and thick while cold.
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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