Answer and Explanation:
(a) The attached image below shows the diagram of a CRT in the helmholtz coils and well labelled with details
(b). The electron will follow a circular path which travels along under constant magnetic field

where m = mass of electron, V = velocity of electron, q = charge of the electron and B = magnetic field strength
The force applied to the cannonball and cannon is equal. The explosion inside the cannon will generate a pressure which will turn into a force on both cannonball and cannon. The cannon being heavier and fixed to the ground will move a bit, but the cannonball will be thrown away, fired.
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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
- the net horizontal force acting on the beam is

where
are the magnitudes of the tensions in ropes 1 and 2, respectively;
- the net vertical force acting on the beam is

where
and
.
Eliminating
, we have





Solve for
.



Explanation:
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