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snow_lady [41]
3 years ago
10

Two small nonconducting spheres have a total charge of 90.0 C.

Physics
1 answer:
valentina_108 [34]3 years ago
8 0

Answer: (a) Smaller charge is 2.7 \times 10^{-5} C and larger charge is 11.7 \times 10^{-5} C.

(b) Smaller charge is -11.4 \times 10^{-5} and larger charge is 9.1 \times 10^{-5}.

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

        Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}

                      = 9 \times 10^{-5} C

Therefore, force between the two spheres will be calculated as follows.

        F = k\frac{Q_{1}Q_{2}}{r^{2}}

       12 N = \frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}

       Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}

or,     Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}

    9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}

    Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0

        Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C

This means that smaller charge is 2.7 \times 10^{-5} C and larger charge is 11.7 \times 10^{-5} C.

(b)  When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

      Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}

      Q_{1} - Q_{2} = 9 \times 10^{-5} C

Now, force between the two spheres is calculated as follows.

    F = k\frac{Q_{1}Q_{2}}{r^{2}}

    12 N = \frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}

   Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}

   Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}

    Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}

        Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}

Hence, smaller charge is -11.4 \times 10^{-5} and larger charge is 9.1 \times 10^{-5}.

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You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you
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Answer:

Approximately 5.9\; {\rm m\cdot s^{-1}} (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

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Overall, the nodes and antinodes on this spring would be:

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The distance between two adjacent nodes is equal to one-half (that is, (1/2)) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, (1/4)) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is \lambda, the length of this spring would be:

\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda.

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The frequency f of this wave is the number of cycles in unit time:

\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}.

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\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}.

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