Answer:

Explanation:
As we know that thermal expansion coefficient of aluminium is given as

now we also know that after thermal expansion the final length is given as

here we know that



now we will have


The second law states that the total entropy can never decrese over time for an isolated system
Answer:
L = 2.8 cm
Explanation:
Period T = 4 / 12 = 1/3 s
T = 2π√(L/g)
L = (T/2π)²g
L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm
From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE =
k 
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE =
k 
10 =
k 
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE =
k 
20 =
(0.2) 
40 = 0.2 
= 200
x = 
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
For more information, check at: brainly.com/question/1352053.