2e level 1
8e level 2
18e level 3
32e level 4
Given:
P1 = 400 kPa
T1 = 110 K
T2 = 235K
Required:
P2
Solution:
Apply Gay-Lussac’s
law where P/T = constant
P1/T1 = P2/T2
P2 = T2P1/T1
P2 = (235K)(400kPa)
/ (110K)
P2 = 855 kPa
Answer:
The elastic potential energy of the spring change during this process is 21.6 J.
Explanation:
Given that,
Spring constant of the spring, 
It extends 6 cm away from its equilibrium position.
We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

So, the elastic potential energy of the spring change during this process is 21.6 J.
False, its up to you to decide what you belive is right and what is wrong.
Answer:
Efficiency of the machine = 75%
Explanation:
Given:
Input work = 8,000 J
Output work = 6,000 J
Find:
Efficiency of the machine
Computation:
Efficiency of the machine = [Output work / Input work]100
Efficiency of the machine = [6,000 / 8,000]100
Efficiency of the machine = 75%