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AleksAgata [21]
3 years ago
10

a macine that increases speed, a machine that changes the direction of force and a machine that increases force

Physics
1 answer:
jasenka [17]3 years ago
7 0
Machine that increases speed, a machine that increases direction of force, and a machine that increases force. A bicycle increases speed, an ax changes direction of force, and a car jack increases force. contrast the scientific definition of work with it's everyday meaning.
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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
pickupchik [31]

Answer:

a

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

b

The focal length of the mirror in water is  f =79.0cm

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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