Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
Answer:
25 mm = 0.984252 inches
Explanation:
Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:
<u>1 mm = 1/25.4 inches</u>
From the question, we have to convert 25 mm into inches
Thus,
<u>25 mm = (1/25.4)*25 inches</u>
So,

Thus, solving we get:
<u>25 mm = 0.984252 inches</u>
The load is placed at distance 0.4 L from the end of
area.
<h3>What is meant by torque?</h3>
The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.
Let the beam is of length L
Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

also, we know that stress at both ends are same


Now from two equations we have

solving the above equation we have

so the load is placed at distance 0.4 L from the end of
area.
The complete question is:
47. the beam is supported by two rods ab and cd that have cross-sectional areas of
and
, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.
To learn more about torque refer to:
brainly.com/question/20691242
#SPJ4
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation

And

a=
b=
Now calculate V1 and V2 at given condition

Substitute given values
= 1 x 10^5 , T = 373.15 and given values of coefficients we get

Solve for V1 by iterative or alternative cubic equation solver we get

Similarly solve for state 2 at P2 = 50 bar we get

Now

a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is

a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series