r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
%d is a format specifier that is a placeholder for an int value. It tells the compiler that we want to print an integer value that is present in variable a. In this way there are several format specifiers in c.
Explanation:
The given data is as follows.
Angular velocity (
) = 2.23 rps
Distance from the center (R) = 0.379 m
First, we will convert revolutions per second into radian per second as follows.
= 2.23 revolutions per second
=
= 14.01 rad/s
Now, tangential speed will be calculated as follows.
Tangential speed, v =
= 0.379 x 14.01
= 5.31 m/s
Thus, we can conclude that the tack's tangential speed is 5.31 m/s.
Answer:
a. E = 122.4 N/C
b. E = 58.2 N/C
c. E = 0
Explanation:
The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.
In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.
A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.