Answer:
More Drag on the down going wing and More Lift on the up going wing
Explanation:
The autorotation spins of blades used in airborne wind energy technology sectors help drive and move the winds and water propeller-type turbines or shafts of generators to produce electricity at altitude and transmit the electricity to earth through conductive tethers.
Sometimes autorotation takes place in rotating parachutes, kite tails. Etc.
As a result, more Drag usually induces the autorotation spin characteristics of a straight-wing aircraft on the downgoing wing and More Lift on the up-going wing.
Answer:

Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80 
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)

The person that is correct based on the 2 statements from Tech A and Tech B is; Tech B
A mass flow sensor is defined as a sensor that is used to measure the mass flow rate of air entering a fuel-injected internal combustion engine and then sends a voltage that represents the airflow to the electronic control circuit.
However, for Tech A is incorrect and so the correct answer is that Tech B is right because his statement corresponds with the definition of mass flow sensor.
Read more about fuel injection engines at; brainly.com/question/4561445
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by


78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa