Answer:
As b ∝ (L/r²) and
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
hence,
the Sun appear brighter in the sky
Explanation:
The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).
thus, mathematically,
b ∝ (L/r²)
now,
given
L for sirius is 23 times more than the sun i.e 23L
now,
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
thus,
using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.
hence, the sun appears brighter
From what I know; When a sample of liquid water vaporizes into water vapor, the electrons in the water sped up due to heat.
5.4*10^-19 C
Explanation:
For the purposes of this question, charges essentially come in packages that are the size of an electron (or proton since they have the same magnitude of charge). The charge on an electron is -1.6*10^-19
Therefore, any object should have a charge that is a multiple of the charge of an electron - It would not make sense to have a charge equivalent to 1.5 electrons since you can't exactly split the electron in half. So the charge of any integer number of electrons can be transferred to another object.
Charge = q(electron)*n(#electrons)
Since 5.4/1.6 = 3.375, we know that it can not be the right answer because the answer is not an integer.
If you divide every other option listed by the charge of an electron, you will get an integer number.
(16*10^-19 C)/(1.6*10^-19C) = 10
(-6.4*10^-19 C)/(1.6*10^-19C) = -4
(4.8*10^-19 C)/(1.6*10^-19C) = 3
(5.4*10^-19 C)/(1.6*10^-19C) = 3.375
(3.2*10^-19C)/(1.6*10^-19C) = 2
etc.
I hope this helps!
Answer:
Explanation:
After the collision velocity of the particle is (4î - 3ĵ)m/s . ... A particle of mass 1 kg moving with a velocity of (4i^−3j^)m/s collides with a fixed surface. ... Perfectly inelastic. D ... The common velocity of the blocks after collision is: ... A ball falls from a height of 5 m and strikes the roof of a lift. ... Stay upto date with our Newsletter! i know this is not right but just here for points see ya loser
