It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
The total angular momentum of the system about point B is
Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.
A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.
Given,
mass of ball 1 = m1
m₂ mass of ball 2=m2
v₁ is the velocity of ball=r₁ω₁
v₂ is the velocity of ball 2=r₂ω₂
The total angular momentum is given as;
Hence the total angular momentum will be
To learn more about angular momentum refer here
brainly.com/question/29512279
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Answer:
pahingi po ng pic pls para masagutang kopo iyan