If an object measures to be 80 centimeters in length, then it is equivalent to ____
1 answer:
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Answer:
The magnetic field strength required to hold anti-protons, moving at 5.70 ✕ 10⁷ m/s in a circular path of 3.20 m in radius is 0.186 T.
Explanation:
To solve the question we note that the magnetic force on a moving charge is given by
F = q·v·B
Where
q = Charge
v = Velocity of the charge =5.70 ✕ 10⁷ m/s
B = Magnetic field
Based on Newton's second law,
Force = Mass, m × Acceleration, a = m × a
Where:
a = Acceleration
m = Mass of anti-proton = Mass of proton = 1.6726219 × 10⁻²⁷ kg
We note that for circular motion, acceleration a is given by
.
Where:
r = Radius = 3.20 m
Therefore, for the circular motion, force, F =
Equating the magnetic force equation to the circular force equation, we have
= q·v·B So that, we find B by making the subject of the formula as follows
. Which gives
= 0.186 T
The magnetic field strength is
B = 0.186 T
To solve this problem we will apply the concepts related to energy conservation. Here we will understand that the potential energy accumulated on the object is equal to the work it has. Therefore the relationship that will allow us to calculate the height will be
Here,
m = mass
g = Acceleration due to gravity
h = Height
our values are,
Replacing,
Then the height is 32.83m.