A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm?

  3.2 J

Step-by-step explanation:

It is convenient to use newtons and meters for the units when calculating work.

The spring constant (k) is ...

  k = (30 N)/(15 cm -12 cm) = (30 N)/(.03 m) = 1000 N/m

For some displacement d, the work done in stretching the spring is ...

  W = (1/2)kd^2 = (1/2)(1000 N/m)(.08 m)^2 = (1/2)(1000)(.0064) Nm

  = 3.2 J

3.2 joules of work is done stretching the spring.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

lol we need to see what the question is asking about

step-by-step explanation: