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:
brainly.com/question/2004529
#SPJ4
I think it was Isaac Newton
I have no idea I need the answer too
Answer:6 joules
Explanation:
Mass(m)=3kg
Velocity(v)=2m/s
Kinetic energy=0.5 x m x v^2
Kinetic energy=0.5 x 3 x 2^2
Kinetic energy=0.5 x 3 x 2 x 2
Kinetic energy=6
Similar elements with similar properties were in the same groups and periods for instance lithium(Li) and sodium(Na) are alkaline metals and so belong to the same group (that is group 1).Also Hydrogen(H) and Helium(He) both have only one shell or energy level and so belong to the same period.