Answer:
g' = 13.5 m/s²
Explanation:
The acceleration due to gravity on surface of earth is given by the formula:
g = GMe/Re² --------------- euation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
Me = Mass of Earth
Re = Radius of Earth
Now, the the acceleration due to gravity on the surface of Kepler-62e is:
g' = GM'/R'² --------------- euation 1
where,
g' = acceleration due to gravity on surface of Kepler-62e
G = Universal Gravitational Constant
M' = Mass of Kepler-62e = 3.57 Me
R' = Radius of Kepler-62e = 1.61 Re
Therefore,
g' = G(3.57 Me)/(1.61 Re)²
g' = 1.38 GMe/Re²
using equation 1:
g' = 1.38 g
where,
g = 9.8 m/s²
Therefore,
g' = 1.38(9.8 m/s²)
<u>g' = 13.5 m/s²</u>
Answer:
<em>a) A positive current will be induced in the coil</em>
Explanation:
Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets<em>, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.</em>
According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. <em>A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil</em>. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.
Answer:
7.77 x 10⁸ Km
Explanation:
given,
Speed of light = 3 x 10⁵ Km/s
= 3 x 10⁸ m/s
1 light hour = 1.079 x 10⁹ Km
now,
0.72 light hour = 0.72 x 1.079 x 10⁹ Km
= 0.7769 x 10⁹ Km
= 7.77 x 10⁸ Km
The Jupiter is 7.77 x 10⁸ Km far from sun.
Answer: 
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity
of the planet P1 with a period
:
(1)
Where:
is the velocity of planet P1
is the radius of the orbit of planet P1
Finding
:
(2)
(3)
(4)
On the other hand, we know the gravitational force
between the star S with mass
and the planet P1 with mass
is:
(5)
Where
is the Gravitational Constant and its value is 
In addition, the centripetal force
exerted on the planet is:
(6)
Assuming this system is in equilibrium:
(7)
Substituting (5) and (6) in (7):
(8)
Finding
:
(9)
(10)
Finally:
(11) This is the mass of the star S
1.) core- the central region where nuclear reactions consume hydrogen to form helium
2.) radiative zone- the section of solar interior between the innermost core and the outer convective zone.
3.) convection zone- where light (photons) is created.
4.) photosphere- the visible surface of the sun that we are most familiar with
5.) chromosphere- second most outer layer of the sun. due to low density it is relatively transparent and only apparent during eclipses.
6.) corona- an aura of plasma that surrounds the sun and other stars but is usually hidden by the bright light of the sun’s surface.