Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.
Answer:
B. Kinetic energy.............
Answer:
dolphin= 7 meters/1 second walrus= 5 meters/1 second
Explanation:
56 divided by 8 is 7
30 diviided by 6 is 5
Using the Hubble law v = H₀d where v = recessional speed = 70,000 km per second H₀ = hubble constant = 70 km/s/Mpc and d = distance of galaxy.
Making d subject of the formula, we have
d = v/H₀
Substituting the values of the variables into the equation, we have
d = v/H₀
d = 70000 km/s/70 km/s/Mpc
d = 1000 Mpc
So, the galaxy is 1000 Mpc away from us.
Learn more about hubble law here:
brainly.com/question/18484687