Answer:transmission
Explanation:
i’m not entirely familiar with this but i’m sure it’s transmission!
Answer:
"150000 N/m²" is the right approach.
Explanation:
According to the question, the pressure on the two spheres 1 and 2 is same.
Sphere 1 and 2:
Then,
⇒ 
⇒ 
and the bulk modulus be,
⇒ 
Sphere 3:
⇒ 
then,
⇒ 
⇒ 
⇒ 
⇒ 
Answer:
First we determine the tensile strength using the equation;
Tₓ (MPa) = 3.45 × HB
{ Tₓ is tensile strength, HB is Brinell hardness = 225 }
therefore
Tₓ = 3.45 × 225
Tₓ = 775 Mpa
From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10
When the percentage of cold work for steel is up to 10,the ductility is 16% EL.
And 16% EL is greater than 12% EL
Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
