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3 years ago

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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What is displacement?

ycow [4]

Answer:

Change in position of an object A vector quantity with unit of distance.

Explanation:

Where is the object going? Final - initial

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2 years ago

5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

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3 years ago

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slavikrds [6]

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3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: