To find the reactance XLXLX_L of an inductor, imagine that a current I(t)=I0sin(ωt)I(t)=I0sin(ωt) , is flowing through the indu
1 answer:
You might be interested in
Answer:
The correct answer is "".
Explanation:
Given:
Maximum load,
P = 50,000 N
Crack length,
a = 3mm
or,
= 3×10⁻³ m
Diameter,
d = 32 mm
As we know,
⇒ Maximum stress,
Now,
⇒ Fracture tougness,
On substituting the values, we get
Answer:
initial diameter of the sample is 2.95 mm
Explanation:
given data
yield load = 2100 N
maximum load = 3400 N
failure load = 2350 N
ultimate engineering stress = 497.4 MPa = 497 × N/m²
to find out
What was the initial diameter of the sample in mm
solution
we will apply here ultimate engineering stress formula that is express as
ultimate engineering stress = ...............1
here A is area and P max is maximum load applied
so area =
here d is initial diameter
so put all value in equation 1
497 × =
solve it we get d
d = 2.95 × m
so initial diameter of the sample is 2.95 mm
Answer:
The phasor technique can't be applied directly in the following cases:
a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)
c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Explanation:
For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.
This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.
This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.
If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:
sin (ωt) = cos (ωt-90º)
This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.
This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).