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kumpel [21]
3 years ago
12

In thinking of an inductor as a circuit element, it is helpful to consider its limiting behavior at high and low frequencies. At

one extreme, the inductor might behave like a short circuit, that is, like a resistor with almost no resistance (an ideal wire) having essentially no voltage drop across it no matter what the current. Alternatively, the inductor might behave like an open circuit, that is, like a resistor with large resistance so that essentially no current will flow no matter what the applied voltage. Based on the formula you obtained for the reactance, how does an inductor behave at high and low frequencies?
A) like a short circuit at both high frequencies and low frequenciesB) like an open circuit at both high frequencies and low frequenciesC) like an open circuit at high frequencies and a short circuit at low frequenciesD) like an open circuit at low frequencies and a short circuit at high frequencies
Physics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

Lyrics C :like an open circuit at high frequencies and a short circuit at low frequencies

Explanation:

The impedance produce by the presence of an inductance in an electric circuit  has a general equation  Z(r) = jωl  where  j is the imaginay number, l is the inductor in henrios and ω is the frecuency of the curent ( ω = 2πf , f is the frecuency in hertz)

So module of the inductance directly depends on  ω = 2πf   then when f the frecuency increase, the value of ωl, also increase and when the value of f decrease ωl will decrease. For very high value of ω the inductance can become very very high ( even up to an open circuit ) and the              opposite for very low values of frecuencies ωl could be very low (as to a short circuit

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GarryVolchara [31]

Power can be defined as the rate at which work is accomplished.

Option D is the correct answer.

<h3></h3><h3>Power </h3>

The work done by an object in a given time interval is called the power of that object.

Suppose an external force F is applied to any object for the time interval T seconds. Due to this external force, the object will perform some amount of work for the time T seconds. This work W done by the object for the time interval T seconds is called the power of that object.

Power can be defined in mathematical term which is given below.

P = \dfrac {W}{T} \;\rm Watts

Thus the power can also be defined as the work done by the object per unit time interval.

Hence we can conclude that option D is the correct answer.

To know more about power, follow the link given below.

brainly.com/question/1618040.

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How much force is needed to stop a 4000 kg truck moving at 8 m/s in 0 2 seconds?​
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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

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Answer:

the same

Explanation:

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