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2 years ago

Which element has more than 5 valency

Physics

1 answer:

Mkey [24]2 years ago

5 0

Answer:

Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.

Explanation:

The electrons found in an element's outermost atomic shell are known as valence electrons.

Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.

However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

Thank you ,

Eddie

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A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} F ⃗ so that the swing ropes

Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

T cos θ = m g

T cos 30° = 30 x 9.8

T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

= 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

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3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

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$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

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$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

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3 years ago

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s344n2d4d5 [400]

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3 years ago

Help! I'm not sure my answer is correct. Problem is attached!

kaheart [24]