(a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 µT. (
1 answer:
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To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.
PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation
Where,
G = Gravitational Universal Constant
d = Distance
M = Mass
Radius earth center of mass
PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,
PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is
At the same time we have that centripetal acceleration is given as
Replacing
Answer:
150000000000 m
0.0000005 seconds
33.33 ns
Explanation:
Speed of electromagnetic waves through vacuum =
Echo time = 1000 seconds
Echo time is the time taken to reach the object and come back to the observer
Distance = Speed×Time
Venus is 150000000000 m away from Earth
Time = Distance / Speed
Echo time will be twice the time
The echo time will be 0.0000005 seconds
Difference in time = Difference in distance / Speed
The accuracy by which I will be able to measure the echo time is 33.33 ns