Answer:
a. mechanical properties
b. thermal properties
c. chemical properties
d. electical properties
e. magnetic properties
Explanation:
a. The mechanical properties of a material are those properties that involve a reaction to an applied load.The most common properties considered are strength, ductility, hardness, impact resistance, and fracture toughness, elasticity, malleability, youngs' modulus etc.
b. Thermal properties such as boiling point , coefficient of thermal expansion , critical temperature , flammability , heat of vaporization , melting point ,thermal conductivity , thermal expansion ,triple point , specific heat capacity
c. Chemical properties such as corrosion resistance , hygroscopy , pH , reactivity , specific internal surface area , surface energy , surface tension
d. electrical properties such as capacitance , dielectric constant , dielectric strength , electrical resistivity and conductivity , electric susceptibility , nernst coefficient (thermoelectric effect) , permittivity etc.
e. magnetic properties such as diamagnetism, hysteresis, magnetostriction , magnetocaloric coefficient , magnetoresistance , permeability , piezomagnetism , pyromagnetic coefficient
As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t. Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.Choose the correct equations of motion for the rotor. (You must provide an answer before moving on to the next part.)
a) a = 2kt, w = 3krº, and 8 = 4kr
b) a = {kt, w = ke?, and 0 = }ke?
c) a = kr?, w = jke', and 0 = krº
d) a = kt, w = jke?, and 0 kr
Answer:
α = kt
ω = [tex]\frac{kt^2}{2}[/tex]
θ = [tex]\frac{kt^3}{6}[/tex]
Explanation:
given data
Initial velocity ω = 0
time t = 10 s
Number of revolutions = 20
solution
we will take here first α = kt .....................1
and as α = [tex]\frac{d\omega}{dt}[/tex]
so that
[tex]\frac{d\omega}{dt}[/tex] = kt ..................2
now we will integrate it then we will get
∫dω = [tex]\int_{0}^{t} kt\ dt[/tex] .......................3
so
ω = [tex]\frac{kt^2}{2}[/tex]
and
ω = [tex]\frac{d\theta}{dt}[/tex] ..............4
so that
[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{kt^2}{2}[/tex]
now we will integrate it then we will get
∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex] ...............5
solve it and we get
θ = [tex]\frac{kt^3}{6}[/tex]