Explanation:
It is given that,
Mass of lithium, 
It is accelerated through a potential difference, V = 224 V
Uniform magnetic field, B = 0.724 T
Applying the conservation of energy as :


q is the charge on an electron

v = 78608.58 m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :



r = 0.0078 meters
So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.
Answer:
finding Cepheid variable and measuring their periods.
Explanation:
This method is called finding Cepheid variable and measuring their periods.
Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.
A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.
Answer:
C. less than 950 N.
Explanation:
Given that
Force in north direction F₁ = 500 N
Force in the northwest F₂ = 450 N
Lets take resultant force R
The angle between force = θ
θ = 45°
The resultant force R


R= 877.89 N
Therefore resultant force is less than 950 N.
C. less than 950 N
Note- When these two force will act in the same direction then the resultant force will be 950 N.
Answer:

Explanation:
M = Mass of each star
T = Time period = 15.5 days
v = Orbital velocity = 230 km/s
G = Gravitational constant = 
Radius of orbit is given by

We have the relation

The mass of each star is 