A good scientific question has certain characteristics. It should have some answers (real answers), should be testable (can be tested by someone through an experiment or measurements), leads to a hypothesis that is falsifiable (means it should generate a hypothesis that can be shown to fail), etc.
Answer:
heterogeneous mixture has components that are not evenly distributed. This means that you can easily distinguish between the different components.
Answer:
b. a lens does not focus all colors of light to the same place.
Explanation:
Chromatic aberration is a defect of a lens. In this defect, the lens is unable to focus the different wavelengths of the light on a single focal point. It is also known as chromatic distortion and color fringing. It is caused by the dispersion of light while passing through a lens. As a result, the image might become blurred and different colors are observed around its edges. It can be corrected by the use of a combination of converging and diverging lenses.
Hence, the correct option will be:
<u>b. a lens does not focus all colors of light to the same place.</u>
A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
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
