Answer:
Explanation:
xx=33 vvalues
Example 1: the two dimensional points P1(0,0) and P2(1,0) and the two tangents P', (1,1) and P2 (0,-1).find the equation of the
1 answer:
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Answer:
The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking
XL=Xc
Substituting the definitions of XL and XC,
2foL=1/2
foC
Solving this expression for fo yields
fo=1/2
where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.
Explanation:
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.
A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.
Answer:
A) while (num >= 0)
Explanation:
To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.
A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.
B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0
C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result
D) The same problem than C
E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result
Answer :
The force needed to move the cylinder is 25.6 N
<h2>Further explanation </h2>Given that,
Length of the cylinder, l = 0.5 m
Outer diameter of the cylinder, d = 8 cm = 0.08 m
Outer radius of the cylinder,
Inside diameter of the pipe, d = 8.5 cm = 0.085 m
Inside radius of the pipe,
Specific gravity of the oil,
Density of oil,
Kinematic viscosity of the oil,
Velocity of the cylinder, u = 1 m/s
We need to find the force needed to move the cylinder. Let the force is F.
Specific gravity is defined as the ratio of the density of the substance to the density of water.
Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.
Absolute viscosity is given by :
Where, = density of oil
And (density of oil = specific gravity × density of water )
So,
..............(1)
The separation between the cylinder and pipe is given by :
are diameter of pipe and cylinder respectively.
The mathematical expression for the Newton's law of viscosity can be written as:
..........(2)
Where
= Shear stress,
............(3)
= viscosity
= rate of shear deformation
On rearranging equation (1), (2) and (3) we get :
...............(4)
A is the area of the cylinder,
Equation (4) becomes :
..............(5)
Now, equation (5) becomes :
F = 25.6 N
<h2>Learn more </h2>Kinematic viscosity : brainly.com/question/12947932
<h2>Keyword : </h2>Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.