To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that
Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by
PART A) Replacing the values to find a, we have
Therefore the semimajor axis is 
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by
Answer: 168.75 N
Explanation:
first, let's convert microcoulombs to coulombs
q1 = 1e-4 C
q2 = 3e-5 C
r = 0.4 m
then use the equation Fe = 
plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2
F = 168.75 N
Answer:
0.48 m
Explanation:
I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others
If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.
0.42 m + 0.06 m = 0.48 m
At the highest point thus, the height is 0.48 m above the ground.
U need to set up n solve the general eqn for simple harmonic motion:
x" = -(k/m)x
solution is x(t) = (x0)*cos(wt) + (v0/w)*sin(wt)
where w=sqrt(k/m), x0 is x-position at t=0 and v0 is vel at t=0
u already calculated f in Q.2 and w = 2*pi*f
x0 is 0 as it starts at eqm
v0 is given at 5.1
so u have x(t)
vel is given by x'(t) = (x0)*(-w)*sin(wt) + (v0/w)*w*cos(wt)
substitute t=0.32, x0=0, v0=5.1 n w in the above, u can solve for v at t=0.32.
