Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding
ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position
Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion
answer✿࿐
I was not able to write it here
so I did it somewhere else and attached the picture
i hope it helps
have a nice day
#Captainpower
Answer:
The force due to air resistance is 256 N.
Explanation:
Given;
mass of the plane, m = 5 kg
applied force on the plane, Fa = 706 N
the net force on the plane, ∑F= 450 N
Let the force due to air resistance = Fr
The net force on the plane is given as;
Net force = applied force - force due to air resistance
∑F = Fa - Fr
Fr = Fa - ∑F
Fr = 706 - 450
Fr = 256 N.
Therefore, the force due to air resistance is 256 N.
