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kari74 [83]
3 years ago
5

A mass is bobbing up and down on a spring. If you increase the amplitude of the motion, how does this affect the time for one os

cillation?
Physics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

It will not change

Explanation:

The period of oscillation of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

T is the period

m is the mass hanging on the spring

k is the spring constant

As we see from the formula, the period of oscillation does not depend on the amplitude of the motion: therefore, if we change the amplitude, the time for one oscillation will not change.

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A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i
mars1129 [50]

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  E = 7500 \ N/C

Explanation:

From the question we are told that  

       The  radius of the inner circle is r_i  =  0.80  \ m

        The  radius of the outer circle is  r_o  =  1.20 \ m

       The  charge on the spherical shell q_n  =  -500nC  = -500*10^{-9} \ C

      The magnitude of the point charge at the center is  q_c =  + 300 nC  =  + 300 * 10^{-9} \ C

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 E =  \frac{k *  q_c   }{x^2}

substituting values  

                  E =  \frac{k *  q_c   }{x^2}

where  k is  the coulomb constant with value k = 9*10^{9}  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

     substituting values

                  E =  \frac{9*10^9  *  300 *10^{-9}}{0.6^2}

                 E = 7500 \ N/C

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