The answer is a no problem
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
Answer: B = 1.69 T
Explanation: Given that the
Electron velocity V = 7.4 × 10^5 m/s
Force F = - 2 × 10^-13 N
Charge q = -1.602 x 10-19 C.
We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation F = qvB sin θ .
Since the electron is moving perpendicular to a magnetic field,
sin θ = sin 90 = 1, therefore
- 2 × 10^-13 = - 1.602×10^-19 × 7.4×10^5B
B = -2×10^-13/1.2×10^-13
B = 1.68708T
B = 1.69 T
Directly outside of the nucleus you have a membrane