A star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears. Therefore, if two stars have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!
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Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer
The pressure of the atmosphere, when a barometer reads 780 mm Hg. Mercury which a density of 1.36 x 10^4 kg /m^3 is B 1.1 x 10^5 N/m^2
This problem can be solved using the formula below
P = dgh................. Equation 1
Where P = Pressure of the atmosphere, d = density of the mercury, h = height of the mercury, g = acceleration due to gravity.
From the question,
Given: d = 1.36×10⁴ kg/m³, h = 780 mm = 0.78 m,
Constant: g = 10 m/s²
Substitute these values into equation 1
P = (1.36×10⁴)(10)(0.78)
P = 10.608×10⁴ N/m²
P ≈ 1.1×10⁵ N/m²
Hence the right answer is B. 1.1×10⁵ N/m²
Learn more about Pressure here: brainly.com/question/23603188
here we have to calculate the net positive charge present on he surface of the conducting sphere.
as the sphere is a conducting one the,the charge will be stored on its surface.
though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.
the electric field at outside of the sphere is E=
as E=
so V=
here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .
hence we have 27=
q=27× coulomb
=3.6× coulomb [ans]