Aforce of 60 n is used to stretch two springs that are initially the same length. spring a has a spring constant of 4 n/m, and spring b has a spring constant of 5 n/m.
how do the lengths of the springs compare?
a: spring b is 1 m longer than spring a because 5 – 4 = 1.
b: spring a is the same length as spring b because 60 – 60 = 0.
c: spring b is 60 m longer than spring a because 300 – 240 = 60.
d: spring a is 3 m longer than spring b because 15 – 12 = 3.

D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Step-by-step explanation:

In this question, you should remember the Hooke's Law in physics.

The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.

The mathematical expression for this law is

where;

F= force applied on the spring

x = the extension on the spring

k= the spring constant which varies in spring.

The question will need you to calculate the extension on the springs A and B then compare the values obtained.

In spring A

Force, F=60N and spring constant ,k=4 N/m

To find the extension x apply the expression;

Here the spring extension is 15 m

In spring B

Force, F=60N and spring constant , k=5N/m

To find the extension x apply the same expression

Here the extension on the spring is 12 m

Compare

The extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m

y=1/4x-4

step-by-step explanation: