Ascale used to weigh fish consists of a spring hung from a support. the spring's equilibrium length is 10.0 cm. when a 4.0 kg fish is suspended from the end of the spring, it stretches to a length of 13.4 cm. (part a) what is the spring constant k for this spring? express your answer with the appropriate units. (part b) if an 8.0 kg fish is suspended from the spring, what will be the length of the spring? express your answer with the appropriate units.

The spring constant of the spring is 1,152.94 N/m.

The new length of the spring when 8kg fish is suspended on it is 16.8 cm.

The given parameters;

The extension of the spring is calculated as follows;

x = L₂ - L₁

x = 13.4 cm - 10.0 cm

x = 3.4 cm

The spring constant of the spring is calculated as follows;

F = kx

mg = kx

The new extension of the spring when 8 kg fish is suspended on it;

The new length of the spring = 6.8 cm + 10 cm = 16.8 cm

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A) 1153 N/m

We can find the spring constant by using Hooke's law:

where

F is the force applied to the spring

k is the spring constant

x is the displacement of the spring

In this problem, a fish of mass m = 4.0 kg is hanging on the spring, so the force applied is the weight of the fish:

and the displacement of the spring is:

so, the spring constant is

B) 16.8 cm

In this case, a fish of mass

m = 8.0 kg

is hanging on the spring. Therefore, the force applied to the spring is

So we can find the displacement of the spring:

And since the equilibrium length of the spring is

the new length of the spring will be

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge