Answer:
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Explanation:
We can answer this question by using Kepler's second law of planetary motion, which states that:
"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"
This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).
In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Reflection: you look in the mirror.
Refraction: You put a straw in a glass of water, and it looks like it broke.
Absorption: If you have a black sweater and you wear it out in the cold, the black sweater is going to hold in heat better than a lighter sweater because the black sweater absorbs light .
Answer:
a). 1.218 m/s
b). R=2.8
Explanation:


Momentum of the motion the first part of the motion have a momentum that is:


The final momentum is the motion before the action so:
a).




b).
kinetic energy

Kinetic energy after

Kinetic energy before

Ratio =

Answer:
Explanation:
<u></u>
<u>1. Formulae:</u>
Where:
- E = kinetic energy of the particle
- λ = de-Broglie wavelength
- m = mass of the particle
- v = speed of the particle
- h = Planck constant
<u><em>2. Reasoning</em></u>
An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.
A proton has mass number 1.
Thus, the relative masses of an alpha particle and a proton are:

For the kinetic energies you find:


Thus:


From de-Broglie equation, λ = h/(mv)

Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J