Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
Answer:
The solved problem is in the photo. Hope it helps.
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:
False
Explanation:
A white dwarf star that is easy to locate and see with small telescopes.