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VashaNatasha [74]
3 years ago
8

When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic

energy?
Physics
2 answers:
Serhud [2]3 years ago
6 0

Answer:

KE : TE = 3 : 4

Explanation:

As we know that the total mechanical energy of the object which is executing SHM is given by

E_{total} = \frac{1}{2}KA^2

here we know that

A = amplitude of SHM

K = spring constant

now we know that total mechanical energy of the spring is always constant so here we can say

kinetic energy + Potential energy = total mechanical energy

we know that potential energy of the spring at any given distance is

U = \frac{1}{2}kx^2

at given position x = A/2

U = \frac{1}{2}K(\frac{A}{2})^2 = \frac{1}{8}KA^2

now we have

KE + \frac{1}{8}KA^2 = \frac{1}{2}KA^2

KE = \frac{3}{8}KA^2

now ratio of kinetic energy and total mechanical energy will be given as

KE : TE = \frac{3}{8}KA^2 : \frac{1}{2}KA^2

KE : TE = 3 : 4

USPshnik [31]3 years ago
5 0
Total energy is a spring:
E = \frac{1}{2} kx^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2

At x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 \\  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 -  \frac{1}{8} ka^2 =  \frac{3}{8} ka^2

The ration:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} =  \frac{3}{4}
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sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

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=>     W = 1.712 *10^{-20 } \  J    

8 0
3 years ago
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34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

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Substitute x=a and R=a

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E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

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