Which of the following results from the fact that skip signals refracted from the ionosphere are elliptically polarized?A. Digital modes are unusable
B. Either vertically or horizontally polarized antennas may be used for transmission or reception
C. FM voice is unusable
D. Both the transmitting and receiving antennas must be of the same polarization
Answers
Answer:
B
Explanation:
The fact that skip signals refracted from the ionosphere are elliptically polarized is a result of either vertically or horizontally polarized antennas may be used for transmission or reception.
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What is the De Broglie wavelength of an electron under 150 V acceleration?
Q1. Basic calculation of the First law (2’) (a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process? (b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?
Young students show a preference for which modality?
Air is compressed in a piston-cylinder device. List three examples of irreversibilities that could occur
Answers
Answer:
(a) 1/L∫Vdt; integral t [0,1]
(b) 1/L∫Vdt; integral t [ 1, infinity]
Explanation:
An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t
The expression for inductor current is given as:
I= 1/L∫Vdt,
where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.
The formula can also be written as:
I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch
Answers
The moment of force at the given slope about point C is 210 ft-lb.
The given parameters:
- Vertical slope, = 2
- Horizontal slope = 3
- Clockwise moment = 330 ft-lb
- Counterclockwise moment = 420 ft-lb
The magnitude of the two moments are in the following simple ratio;
330:420 = 11:14 (divide through by 30)
- the A coordinate = (0, 5)
- the B-coordinate = (5,0)
The line of action of the force passes line AB at the final following coordinates;
total ratio of 11:14 = 11 + 14 = 25
The position of C = (3, 1)
The resultant position of point C = (3 - 2.2, 2.8-1) = (0.8, 1.8)
The moment of force at the given slope about point C is calculated as;
3(1.8) + 2(0.8) = 7
Recall that this is the simplest form of the moment produced by the force.
Moment about C = 7 x 30 = 210 ft-lb
Thus, the moment of force at the given slope about point C is 210 ft-lb.
Learn more about moment of force here: brainly.com/question/6278006
Answers
Answers
Answer:
891.027 lbm/ft³
Explanation:
See it in the pic
Answers
Answer and Explanation:
The main objective that the torsion test serves is the determination of the material behavior or the behavior of the test sample when subjected to torsional stresses or forces due to the application of moments that results in shear stress along the axis.
Plasticity is the property of elastic material and tension or shear stresses leads to plasticity in a material where these links are the weakest, that gives torsion test a major advantage over the tension test.
Torsion tests are performed on materials to deduct properties like the shear modulus of elasticity, the torsional strength, and the MOR, i.e., Modulus of Rupture.
This test can be used to obtain larger strain values of strain without any complexity as that in tension test.
This test provides a curve of shear-stress-shear strain which is more significant in determining the plasticity as compared to the curve of stress-strain in tension test.
Maximum torque for a given value of maximum stress will be 2 times higher in torsion as that of tension.
In torsion, for plastic flow, the threshold value of shear stress is achieved before the threshold value of normal stress for fracture whereas in tension the critical value of normal stress is achieved sooner than the critical shear for plastic flow.