Technician B is correct because torque is a force of an object.
Answer:
yes it is certainly good ice cream
Answer:
The speed of shaft is 1891.62 RPM.
Explanation:
given that
Amplitude A= 0.15 mm
Acceleration = 0.6 g
So
we can say that acceleration= 0.6 x 9.81

We know that

So now by putting the values



We know that
ω= 2πN/60
198.0=2πN/60
N=1891.62 RPM
So the speed of shaft is 1891.62 RPM.
Answer:
B. G = 333 mS, B = j250 mS
Explanation:
impedance of a circuit element is Z = (3 + j4) Ω
The general equation for impedance
Z = (R + jX) Ω
where
R = resistance in ohm
X = reactance
R = 3Ω X = 4Ω
Conductance = 1/R while Susceptance = 1/X
Conductance = 1/3 = 0.333S
= 333 mS
Susceptance = 1/4 = 0.25S
= 250mS
The right option is B. G = 333 mS, B = j250 mS
Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN