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andrey2020 [161]
3 years ago
14

Can a small force exert a greater torque than a larger force.Explain.

Physics
1 answer:
butalik [34]3 years ago
6 0

Answer:

Yes, a small force can exert  a greater torque than a large if small force has large enough lever arm.

Explanation:

We know that torque exert on an object is given by

\tau=Force\times perpendicular\;distance

Torque is directly proportional to force

\tau\propto F

Torque is directly proportional to perpendicular distance

If force is small then torque exert is small

If force is large then torque exert is large

If lever arm is small then torque is small

If lever arm is large then torque is large.

A small force exert a greater torque than a large force if small force has large enough lever arm.

Yes, a small force can exert  a greater torque than a large.

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The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th
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Answer:

a

The focal length of the lens in water is  f_{water} = 262.68 cm

b

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Explanation:

From the question we are told that

    The index of refraction of the lens material = n_2

    The index of refraction of the medium surrounding the lens = n_1

 

The lens maker's formula is mathematically represented as

            \frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

Where f is the focal length

            n is the index of refraction

            R_1 and R_2 are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           \frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

    When immersed in liquid the formula becomes

          \frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2}  ]

The ratio of the focal length of the the two medium is mathematically evaluated as

           \frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }

From the question

      f_{air }= 79.0 cm

       n_2 = 1.55

and the refractive index of water(material surrounding the lens) has a constant value of  n_1 = 1.33

         \frac{f_{water}}{79}  = \frac{1.55- 1}{\frac{1.55}{1.44}  -1}

           f_{water} = 262.68 cm

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

   

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