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Jet001 [13]
3 years ago
8

Pls,help ASAP .the question is in attachment​

Physics
1 answer:
Alex777 [14]3 years ago
8 0

Answer:c

Explanation:

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A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the r
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Solution

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Correct option is

C

3 cm

RI=apparent depthreal depth

Substituting, 34=apparentdepth12

Therefore, apparent depth=412×3=9

The height by which it appears to be raised is 12−9=3cm

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SIMILAR QUESTIONS

A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because

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>

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

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in the diagram, the solid eventually becomes a gas. what will happen to bring the substance from a solid to a gas?
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3 years ago
Two small nonconducting spheres have a total charge of 90.0 C.
valentina_108 [34]

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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