What is the amount of work done when JoAnne throws a baseball 2 meters at a force of 40
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Answer:
The length of the flagpole is approximately 87.43 m
Explanation:
The given parameters of the cable attached to the flagpole are;
The point along the flagpole at which the cable is attached = 32.0 m
The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°
The downward force exerted by the cable, = 1.233 × 10⁴ N
The force exerted by the cable to the left = 1.233 × 10⁴ N
Let 'W' represent the weight of the flagpole, at equilibrium, we have;
The sum of vertical forces = 0
Therefore;
+ W - R = 0
W - R = -1.233 × 10⁴ N
Taking moment about the support at the base of the pole, we get;
1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0
∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d
W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N
R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373
R ≈ 30,382.373 N
Taking moment about the point of attachment of the cable to the ground, we have;
W × ((d/2) × cos(60.0°) + 32) = R × 32
∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281
d = 2 × 43.71281 ≈ 87.43
The length of the flagpole, d ≈ 87.43 m
The additional charge transferred to the capacitors by the battery is 60 μC.
The increase in the total charge stored on the capacitors is 60 μC.
<u>Explanation</u>:
C = εоA / d
If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.
- Initial charge of the capacitor is:
q = CV
= (6e - 6) (10)
= 60 μC
Final charge of the capacitor:
q = (2C)V
= (2 6e - 6)
(10)
= 120 μC
additional charge transmitted is:
q' = 120 - 60
= 60 μC
- initial total charge:
= (C1 + C2) V
= (6 + 6) (10)
= 120 μF
final total charge:
= (C1 + C2) V
= (2 6 + 6)
(10)
= 180 μF
Increase in the charge:
q' = 180 - 120
= 60 μC