Given:
Wall's inside temperature,
Room air temperature,
Solution:
To calculate percentage max relative humidity, we make use of steam table for saturated pressure of wall and air:
From steam table:
At :
At :
Now,
(RH)_{max} = 68.58%
Therefore, max Relative Humidity of the air before the occurrence of condensation in wall is 68.85%
Answer:288 pm
Explanation:
Number of atoms(s) for face centered unit cell -
Lattice points: at corners and face centers of unit cell.
For face centered cubic (FCC), z=4.
- whereas
For an FCC lattices √2a =4r =2d
Therefore d = a/√2a = 408pm/√2a= 288pm
I think with this step by step procedure the, the answer was clearly stated.
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v =
in this expression the densities are constant
v = A √h
A =
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A =
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h =
Δh = Δv
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
= 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
= 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
Answer:
d. all of the above
Explanation:
There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.
The radial acceleration is given by;
Where;
V is the velocity of the particle
R is the radius of the circular path
This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.
Therefore, from the given options in the question, all the options are correct.
d. all of the above
Answer:
Explanation:
Given data in question
mean stress = 50 MPa
amplitude stress = 225 MPa
to find out
maximum stress, stress ratio, magnitude of the stress range.
solution
we will find first maximum stress and minimum stress
and stress will be sum of (maximum +minimum stress) / 2
so for stress 50 MPa and 225 MPa
=
+
/ 2
50 = +
/ 2 ...........1
and
225 = +
/ 2 ...........2
from eqution 1 and 2 we get maximum and minimum stress
= 275 MPa ............3
and = -175 MPa ............4
In 2nd part we stress ratio is will compute by ratio of equation 3 and 4
we get ratio = /
ratio = -175 / 227
ratio = -0.64
now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.
magnitude = -
magnitude = 275 - (-175) = 450 MPa