Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '
' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.
Answer:
(a) V1 = 8990.00 V
V2 = 8960.13 V
Explanation:
Parameters given:
q =3 mC
k = 8.99 * 10⁹ Nm²/C²
x1 = 3 m
x2 = 3.01 m
Electric potential is given as:
V = kq/r
Where
k = Coulombs constant
q = charge
r = distance
Potential at x1 is:
V1 = (8.99 * 10⁹ * 0.000003)/(3)
V1 = 8990.00V
Potential at x2 is:
V2 = (8.99 * 10⁹ * 0.003)/(3.01)
V2 = 8960.13 V
They travel at different speeds
C. creates biases
this is the only answer that would be a disadvantage.
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