Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:

ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂

θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂

=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain

=3.335 *10^-4

ε(avg) =150 *10^-6
orientation of γmax

θ = 31.71 or -58.29
To determine the direction of γmax

= 1.67 *10^-4
Answer:
The correct option is;
Materials and Components
Explanation:
The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on
1) Advances in fluid power
2) Making use of the advantages
3) Making use of the other externally available technological advantages
4) Giving allowance for disadvantages
Areas of interest in advances in fluid power are;
a. Computer optimized flow
b. The use of new and improved materials/coatings
c. The use of components that save energy, such as intelligent supply pressure adapting systems
Answer:
what
Explanation:
is this an exam or an test or what is it
Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
The response to whether the statements made by both technicians are correct is that;
D: Neither Technician A nor Technician B are correct.
<h3>Radio Antennas</h3>
In radios, antennas are the means by which signals to the sought frequency be it AM or FM are received.
Now, if the antenna is bad, it means it cannot pick any radio frequency at all and so Technician A is wrong.
Now, most commercial antennas usually come around a resistance of 60 ohms and so it is not required for a good antenna to have as much as 500 ohms resistance and so Technician B is wrong.
Read more about Antennas at; brainly.com/question/25789224