Answer:
3.0 x10^-3 J
Explanation:
The potential energy of a spring is given by PE = (0.5)k*x^2
Where
K: Spring Constant = 60 N/m
x: displacement of the spring from its equilibrium position = 1cm = 0.01m
Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J
Translate please, i’d be able to help better:)
Answer:
ρ = 830.32 kg/m³
Explanation:
Given that
Oil head = 12.2 m
h= 12.2 m
Pressure P = 1.013 x 10⁵ Pa
Lets take density of the liquid =ρ
The pressure due to liquid P given as
P = ρ g h
Now by putting the all values in the above equation
1.013 x 10⁵ Pa = ρ x 10 x 12.2 ( take g =10 m/s²)
ρ = 830.32 kg/m³
Therefore the density of oil is 830.32 kg/m³
