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Novay_Z [31]
3 years ago
8

a force of 5 N extends a spring of natural length 0.5m by 0.01, what will be the length of the spring when the applied force is

20 N​
Physics
1 answer:
vova2212 [387]3 years ago
5 0

Answer:

L_new =L+x^2  = L_new = 0.54_m.

Explanation:

Given data:

Force in the first case,  

F_1 = 5N

Force in the second case,  

F_2 = 20 N

Natural length of spring,  

L= 0.5

Extension in the first case,  

x_1 = 0.01m

Let the force constant of the spring be k.

Thus,

F_1=kx_1

5 = k × 0.01  

⇒ k = 500 N/m.

The extension in the spring in the second case can be given as,

F_2=kx_2

20 = 500x_2

⇒ x_2 = 0.04 m.

Thus, the effective length of the spring would be,

L_new =L+x^2

L_new = 0.5+0.04

L_new = 0.54_m.

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Answer:

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Given that,

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v=\dfrac{3\times10^{8}}{1.41}

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Using formula of distance

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a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

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