Which of the following is not a component of skill related fitness

Put the pairs of atoms in order, with the pair that has the biggest electronegativity difference between the two atoms at the to

mojhsa [17]

Electronegativity is the measure of the tendency of an atom to attract a bonding pair of electrons. In the periodic table, electronegativity increase across the period because the charges on the nucleus increase. The correct arrangement for the atoms given above is as follows
Flourine and Francium
Chlorine and Cesium
Nitrogen and Sodium
Phosphorus and Lithium
Nitrogen and Sulphur.

5 0

3 years ago

The equation v=F^aL^M^-c were shows the relationship between velocity of the waves tensile force in the string length, L and mas

Travka [436]

I literally looked everywhere for the answer, and I still found nothing. I hope you get it right. Sorry.

8 0

3 years ago

Urgente!!!!! Necesito ayuda con esto!!!!

ahrayia [7]

media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4

5 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

9) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen

xeze [42]

Answer:

See the answer below

Explanation:

The best thing one can do in this case would be to return the microscope's objective to low power and then re-center the specimen before switching back to high-dry power.

Most of the time, what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage. Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.

After doing the above and the specimen still does not come into focus, then the instructor can be called upon.

4 0

3 years ago