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Answer:
work which is required to move an electron from the positive terminal to the negative terminal is
-4μJ
W = -4μJ
Explanation:
Work required to move the charge
W = qΔV
initial point V₁ = 150V
Final point V₂ = -50V
W = q(V₂ -V₁)
= 20 × 10⁹(-50 - 150)
W = -4μJ
work which is required to move an electron from the positive terminal to the negative terminal is
-4μJ
Answer:
μb = 0.096
μc = 0.073
Explanation:
member AB:
-800( 4/3 ) + Nb (2) = 0
Nb (2) = 3200/3
Nb = 533.3N
Post BC:
summation of force along the y axis=0
Nc + Nb + 150(3/5 ) -50(9.81)=0
Nc + 533.3 + 150(3/5 ) -50(9.81)=0
Nc = 933.83 N
Also (-4/5)(150)(3) + Fb(0.7)= 0
Fb = (4/5)(150)(3)/0.7 = 51.429 N
Likewise alog the x axis,
4/5(150) - Fc -Fb = 0
4/5(150) - Fc -51.429 = 0
Fc = 4/5(150) -51.429 =68.571 N
μb = Fb/Nb = 51.429/533.3 = 0.096
μc = Fc/Nc = 68.571 / 933.83 = 0.073