If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N

Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
= 9.8°
Explanation:
Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.
width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m
angular position of fringe, Sinθ = n λ /a
n is order of fringe , λ is wave length of light and a is slit aperture
So Sinθ ∝ 1 / a
Sin θ₁ /Sin θ₂ = a₂/a₁ ;
Sin20°/sinθ₂ = 2 / 1
sinθ₂ = Sin 20° / 2 = .342/2 = .171
θ₂ = 9.8 °
Advances in technology used to study and observe atoms lead to the discovery of electrons, protons, nuetrons, and the quarq
Answer:
4/3 m/s
Explanation:
Assuming momentum is conserved, the sum of products of mass and speed before the collision is the same as after:
(2000 kg)(4 m/s) +(4000 kg)(0 m/s) = (2000 +4000 kg)(Vt)
Vt = (8000 kg·m/s)/(6000 kg) = 4/3 m/s
The speed of the combined objects after the collision is 4/3 m/s.