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blagie [28]
3 years ago
7

A net force of magnitude 36 N gives an object of mass m1 an acceleration of 6.0 m/s^2. The same net force gives m1 and another o

bject of mass m2 fastened together an acceleration of 2.0 m/s^2. What acceleration will m2 experience if the same net force acts on it alone?
Physics
1 answer:
Jet001 [13]3 years ago
3 0
This is a beautiful problem to test whether a student actually understands
Newton's 2nd law of motion . . . Force = (mass) x (acceleration). 

That simple law is all you need to solve this problem, but you need to
use it a few times.

m₁ alone:
                     Force = (mass) x (acceleration)

                       36 N = ( m₁ ) x  (6 m/s²)

                         m₁ = (36 N) / (6 m/s²)   

                         m₁  =  6 kilograms .

m₁ and m₂ glued together:
                     Force = (mass) x (acceleration)

                       36 N = (6 kg + m₂) x (2 m/s²)

                     6 kg + m₂  =  (36 N) / (2 m/s²)  =  18 kilograms

                               m₂  =  12 kilograms .

m₂ alone:
                     Force = (mass) x (acceleration)

                       36 N = (12 kg) x (acceleration)

                     Acceleration = (36 N) / (12 kg)

                     Acceleration  =   3 m/s²


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Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

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          Image distance would be  v =  -15 \ cm

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              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

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