Answer:
<em>-2 units of charge</em>
Explanation:
charge on A = Qa = -6 units
charge on B = Qb = 2 units
if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.
charge on each sphere will be = ![\frac{Qa + Qb}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BQa%20%2B%20Qb%7D%7B2%7D)
charge on each sphere =
=
= <em>-2 units of charge</em>
Answer
given,
F₁ = 15 lb
F₂ = 8 lb
θ₁ = 45°
θ₂ = 25°
Assuming the question's diagram is attached below.
now,
computing the horizontal component of the forces.
F_h = F₁ cos θ₁ - F₂ cos θ₂
F_h = 15 cos 45° - 8 cos 25°
F_h = 3.36 lb
now, vertical component of the forces
F_v = F₁ sin θ₁ + F₂ sin θ₂
F_v = 15 sin 45° + 8 sin 25°
F_v = 13.98 lb
resultant force would be equal to
![F = \sqrt{F_h^2+F_v^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7BF_h%5E2%2BF_v%5E2%7D)
![F = \sqrt{3.36^2+13.98^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7B3.36%5E2%2B13.98%5E2%7D)
F = 14.38 lb
the magnitude of resultant force is equal to 14.38 lb
direction of forces
![\theta =tan^{-1}(\dfrac{F_v}{F_h})](https://tex.z-dn.net/?f=%5Ctheta%20%3Dtan%5E%7B-1%7D%28%5Cdfrac%7BF_v%7D%7BF_h%7D%29)
![\theta =tan^{-1}(\dfrac{13.98}{3.36})](https://tex.z-dn.net/?f=%5Ctheta%20%3Dtan%5E%7B-1%7D%28%5Cdfrac%7B13.98%7D%7B3.36%7D%29)
θ = 76.48°
The work done to accelerate the acrobat is given by
![W=Fd](https://tex.z-dn.net/?f=W%3DFd)
where F is the force applied and d the distance of application of the force.
If the barrel is 3.05 m long, then d=3.05 m. Therefore we can find the force: