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bulgar [2K]
3 years ago
15

Which of the following is a common property of Noble Gases?

Physics
2 answers:
Softa [21]3 years ago
4 0
<span>Have a full outer shell of electrons.</span>
Ne4ueva [31]3 years ago
3 0

Answer: the property of noble gases which is common is that they have a full outer shell of electrons.

Explanation:

Noble gases are the gases belonging to Group 18 of the periodic table. The electronic configuration for these gases are: ns^2np^6

From the electronic configuration, it is visible that they have fully filled outer shell and hence, these gases are least reactive of all the elements in the periodic table.

As, they are not reactive and therefore they do not form any chemical compounds easily.

From the electronic configuration, it is visible that there are no free electrons in these gases, therefore, they are not good conductors of electricity.

Hence, the correct answer is that they all have full outer shell of electrons.

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A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.
snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  v =  3.79 *10^{5} \ m/s  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  E  =  144 \ kV /m   =  144*10^{3} \  V/m

     The magnetic field is  B  =  0.38 \ T

   

The force due to the electric field is mathematically represented as

      F_e =  E  * q

and

The force due to the magnetic field is mathematically represented as

    F_b  =  q * v  *  B * sin(\theta )

Now given that it is perpendicular ,  \theta  = 90

=>   F_b  =  q * v  *  B * sin(90)

=>   F_b  =  q * v  *  B

Now  given that it is not deflected it means that

        F_ e  =  F_b

=>    q *  E = q *  v  * B

=>   v =  \frac{E}{B }

 substituting values

     v =  \frac{ 144 *10^{3}}{0.38 }

     v =  3.79 *10^{5} \ m/s

7 0
3 years ago
Describe the relationship between air time and range.
kifflom [539]

Answer: For the angles below 40, The air time and range are directly proportional meaning that if air time increases, the range increases. Like said earlier, the total traveled distance in the x-direction depends on the initial velocity and time.

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Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magnetic field. Explain whe
Otrada [13]

Answer:

No.

Explanation:

  • According to Faraday's law, the induced emf in the circuit is given by :

         e=\dfrac{d\phi}{dt}, it is proportional to the rate of change of magnetic flux.

  • In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
  • Hence, no current will induce.
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