Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet, 
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

M is the mass of the sun

T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
Answer:
The convection process plays an important role in the liquid. Due to the increasing heat supply or high amount of temperature, the fluid gets heated up, as a result of which it becomes warm, less dense and eventually rises up forming convection cells.
In the interior of the earth, the hot molten rocks get heated up due to the heat supplied by the core of the earth. This makes the magma warm and less dense and rises upward forming convection currents in the mantle.
This convection process is similar to the convection cells that form in the atmosphere, where the hot, less dense air rises up in the atmosphere forming a low-pressure zone. This uprising air forms convection cells, in which the warm air rises and as it rises high in the atmosphere, the temperature becomes low, making the air cold and it eventually sinks.
Answer:
a. 7.046 Nm²/C
b. 2.348 Nm²/C
Explanation:
Data given:
Base of equilateral triangle = 25.0 cm = 0.25 m
Strength of electric field = 260 N/C
In order to find the electric flux we first have to find out the area of triangle.
Area of triangle = 
= 
= 0.0271 m³
Lets find electric flux,
Electric Flux = E. A
= 260×0.0271
= 7.046 Nm²/C
Now we can find the electric flux through each of the three sides.
Electric flux through three sides = 
= 2.348 N m²/C
The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
<u>Explanation</u>:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v