Answer:
What is the difference Plastic vs elastic deformation
Explanation:
The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.
The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.
Answer:
I don't know ☺️☺️☺️❌‼️
Explanation:
I don't understand this question
Answer: 1766.667 Ω = 1.767kΩ
Explanation:
V=iR
where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).
3mA = 0.003 A
Rearranging the equation, we get
R=V/i
Now we are solving for resistance. Plug in 0.003 A and 5.3 V.
R = 5.3 / 0.003
= 1766.6667 Ω
= 1.7666667 kΩ
The 6s are repeating so round off to whichever value you need for exactness.
Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
