The gas planets usually have extremely high gravitational pulls, the surface isn't solid (since its a gas planet), and gas planets are larger than the inner planets.
<span>Similarities- These planets all have moons and they both revolve around the sun (obviously).
Hope this helps.</span>
Answer:
False, Sunspots appear dark (in visible light) due to their low temperature(cooler) than rest of the sun
Explanation:
Sunspots appear dark because they are much cooler( have low temperature than the rest of the surface contained by Sun. As they appear dark, but still they have very temperature that's why so hot. Sunspots have temperatures ranges 3,500 Celsius (3773 kelvin) and the surrounding surface of the sun has a temperature much higher of about 5,500 Celsius(5773 Kelvin). Even if we see a sunspot alone in space, it will glow so brightly.
Learn more about sunspots :
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Answer:
I = 30 A.
Explanation:
Given that,
The voltage of the battery, V = 230 V
Power used to charge the battery, P = 6.9 kW
We need to find the current used to charge the battery. The formula for the power is given by :
P = VI
Where
I is current
So,
So, the required current is 30 A.
Answer:
0.0025116weber/m²
Explanation:
Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).
Mathematically;
B = ¶/A
¶ = BA
Given B = 0.23Tesla which is the magnitude of the magnetic field
Dimension of the rectangular loop = 7.8 cm by 14 cm
Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm
= 109.2cm²
Converting this value to m²
Area of the loop = 109.2 × 10^-4
Area of the loop = 0.01092m²
Magneto flux = 0.23×0.01092
Magnetic flux = 0.0025116weber/m²
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4
