Answer:
V_vap = 161.2 L
Explanation:
The total mass of the aluminum rod is given as;
m = ρ∙V = ρ∙L∙A
Where;
ρ is density = 2700 kg/m³
L is length = 3.3m
A is cross sectional area = 3.8 cm² = 3.8 x 10⁻⁴ m²
Thus;
m = 2700kg/m³•3.3m•3.8 × 10⁻⁴m²
= 3.3858kg
By cooling down the submerged half of the aluminum rod releases an heat amount of
Q = (1/2)∙m∙cp∙∆T
Where;
cp is specific heat of aluminum aluminum = 900 J/kg
∆T is change in temperature = 274 - 4.2 = 269.8 K
Thus;
Q = (1/2)•3.3858•900•(269.8)
= 411069.978 J
The liquid absorbs this heat and vaporizes partially, such that the heat equals vaporized mass times latent heat of vaporization:
Q = m_vap•∆h_vap
Making m_vap the subject;
m_vap∙ = Q/∆h_vap
Where ∆h_vap is latent heat of vaporization given as 20900J/kg
Thus,
m_vap∙ = 411069.978/20900
= 19.668 kg
Let's divide this mass by the density of liquid helium and we get the liquid volume which has vaporized:
V_vap∙= m_vap/ρ
V_vap∙ = 19.668/122
V_vap∙ = 0.1612 m³
Converting to litres;
V_vap = 0.1612 x 1000
V_vap = 161.2 L
The answer is hydrogen ions
To solve this problem it is necessary to apply the concepts related to optical magnification (is the process of enlarging the apparent size, not physical size, of something.). Specifically the angular magnification of an optical telescope is given by
Where,
Focal length of the objective lens in a refractor
Focal length of the eyepiece
Our values are given as
71cm
2.1cm
Replacing we have
Therefore the magnification of this astronomical telescope is -33.81
Answer:
201.6 N
Explanation:
m = mass of disk shaped merry-go-round = 125 kg
r = radius of the disk = 1.50 m
w₀ = Initial angular speed = 0 rad/s
w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s
t = time interval = 2 s
α = Angular acceleration
Using the equation
w = w₀ + α t
4.296 = 0 + 2α
α = 2.15 rad/s²
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²
F = constant force applied
Torque equation for the merry-go-round is given as
r F = I α
(1.50) F = (140.625) (2.15)
F = 201.6 N
That's wave 'diffraction'.
