Answer:
38 m/s
43 m/s
Explanation:
x = 18t + 5.0t²
The instantaneous velocity is the first derivative:
v = 18 + 10.t
At t = 2.0:
v = 18 + 10.(2.0)
v = 38 m/s
The average velocity is the change in position over change in time.
v = Δx / Δt
v = [ (18t₂ + 5.0t₂²) − (18t₁ + 5.0t₁²) ] / (t₂ − t₁)
Between t = 2.0 and t = 3.0:
v = [ (18(3.0) + 5.0(3.0)²) − (18(2.0) + 5.0(2.0)²) ] / (3.0 − 2.0)
v = [ (54 + 45) − (36 + 20.) ] / 1.0
v = 99 − 56
v = 43 m/s
Shear stress = 1.0 N/m² (Pa)
For water, the dynamic viscosity = 10⁻³ Pa-s at 20°C.
The velocity gradient required = (Shear stress)/(Dynamic viscosity)
= (1.0 Pa)/( 10⁻³ Pa-s)
= 10³ 1/s
Answer: 10³ s⁻¹
The answer would be A) Ductility because the definition of ductile means that something is able to be pulled through a wire (i.e: copper)