Answer:
1.5 m
Explanation:
Let the distance from the box to the pivot be c.
Let the distance from the pivot to the effort be y.
From the question given above, the following data were obtained:
Effort force (Fₑ) = 7 N
Force of resistance (Fᵣ) = 14 N
Distance from the box to the pivot (c) = 0.75 m
Distance from the pivot to the effort (y) =?
Clockwise moment = Fₑ × y
Anticlock wise moment = Fᵣ × c
Clockwise moment = Anticlock wise moment
Fₑ × y = Fᵣ × c
7 × y = 14 × 0.75
7 × y = 10.5
Divide both side by 7
y = 10.5 / 7
y = 1.5 m
Therefore, the distance from the pivot to the effort is 1.5 m
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
