Normal force is mass x gravity, so mass x 9.81
In almost every case in nature, adding heat to a liquid
causes the density of the liquid to decrease. That is,
when the liquid gets warmer, it expands and occupies
more space.
The one big exception to this rule is water !
Starting with a block of ice at zero°C (32°F), as the ice melts,
becomes water at zero°C, and all the way to 4°C (about 39°F),
its density increases all the way. That is, it shrinks and occupies
less volume as it goes from ice at zero°C to water at 4°C.
This sounds like an interesting but insignificant quirk ... until
you realize that if water didn't do this, then life on Earth would
be impossible !
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
Answer:
The motion is over-damped when λ^2 - w^2 > 0 or when
> 0.86
The motion is critically when λ^2 - w^2 = 0 or when
= 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when
< 0.86
Explanation:
Using the newton second law
k is the spring constante
b positive damping constant
m mass attached
x(t) is the displacement from the equilibrium position

Converting units of weights in units of mass (equation of motion)

From hook's law we can calculate the spring constant k

If we put m and k into the DE, we get

Denoting the constants
2λ =
= 
λ = b/0.215

λ^2 - w^2 = 
This way,
The motion is over-damped when λ^2 - w^2 > 0 or when
> 0.86
The motion is critically when λ^2 - w^2 = 0 or when
= 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when
< 0.86