Most of the stars in our universe are in binary systems. Hence, the size of the stars can be found when one star eclipses the other. When so happens, the change in the luminosity is measured and size is calculated. Another way is by measuring the luminosity of the star and comparing it with the Sun's luminosity. As the luminosity is dependent on the size of the star, the radius of the star can be calculated when compared to the Sun. Following formula can be used:
Star's radius/Sun's radius = (Sun's temperature/star's temperature)2 Sqrt[star's luminosity/Sun's luminosity].
Answer:
Acceleration of the proton will be equal to 
Explanation:
We have given electric field 
Mass of proton is equal to 
And charge on proton is equal to 
Electrostatic force will be responsible for the motion of proton
Electrostatic force will be equal to 
According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration
This newton force will be equal to electrostatic force
So 
So acceleration of the proton will be equal to
Answer: An object undergoing uniform circular motion is moving
Explanation:
Here, we are required to determine which combination of molecules will produce ammonia with no leftovers.
Option A: 2N2 and 6H2 is the correct combination of molecules that will produce ammonia with no leftovers.
First, it is important to know that both Nitrogen and Hydrogen used in the production of ammonia are diatomic.
Secondly, Nitrogen and Hydrogen are in the ratio 1 : 3.....
As such, the coefficient of hydrogen should be thrice that of Nitrogen to ensure that there are no leftovers.
Therefore, option A which has:
2N2 and 6H2 is the correct combination of molecules that will produce ammonia with no leftovers.
Read more:
brainly.com/question/24396848
It sounds like a special relativity question but I need more info for a total answer. But remember it's length in the lab frame is
L•sqrt(1-(v/c)^2) where L is the rest length, v is its velocity magnitude and c is the speed of light. Sqrt is the square root (I'm on a phone so I can't see the math equation editor)
