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postnew [5]
3 years ago
10

10. In deep space, two spheres of negligible size are connected by a 200-meter nonconducting cord. If a uniformly distributed ch

arge of 0.04 C resides on the surface of each sphere, calculate the tension in the cord. a. 1,800 newtons b. 9,600 newtons c. 18,000 newtons d. 4,500 newtons e. 0 newtons
Physics
1 answer:
tigry1 [53]3 years ago
6 0

Answer:

359.2 N

Explanation:

The tension in the cord is due to the repulsive electrostatic force between the charged spheres connected by the cord.

T=k\frac{Q^2}{d^2}

k is the Coulomb constant = 8.98×10⁹ Nm²/C²

Q= 0.04 C

d = 200 m

T=8.98\times10^9\frac{(0.04)^2}{(200)^2} =359.2 N

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Explanation:

The given data is as follows.

     m_{1} = 57 kg,        m_{2} = 79 kg

      l_{1} = 6.5 m,         l_{2} = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1} = 0

     57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5 = 0

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Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}    

  57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5 = 0

                            T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.  

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