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elena-14-01-66 [18.8K]

3 years ago

9

What is the relationship between the valence electrons of an atom and the chemical bonds the atom can form?

1 answer:

stellarik [79]3 years ago

8 0

Answer:

Valence electrons are outer shell electrons with an atom and can participate in the formation of chemical bonds. In single covalent bonds, typically both atoms in the bond contribute one valence electron in order to form a shared pair. The ground state of an atom is the lowest energy state of the atom.

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the magnitude of the normal force acting on a person with mass of 70 kg standing at rest on the flat ground would be ?

Mekhanik [1.2K]

Answer:

$f = mg \\ = 70 \times 9.8 = |f|$

3 0

2 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5

juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

or

$T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}$

Where

l is length of the pendulum

$l\propto T^2$

or

$\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2$ ....(1)

ATQ,

$\dfrac{T_1}{T_2}=1.5$

Put in equation (1)

$\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}$

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0

2 years ago

You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal

RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

$K_{total} = K_{traslation} + K_{rotation}$

let's look for each energy

linear

$K_{traslation}$ = ½ m v²

rotation

$K_{rotation}$ = ½ I w²

the moment of inertia of a solid sphere is

I = 2/5 m r²

we substitute

$K_{total}$ = ½ mv² + ½ I w²

angular and linear velocity are related

v = w r

we substitute

K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

K_{total} = m w² r² (½ + 1/5)

K_{total} = $\frac{7}{10}$ m w² r²

let's calculate

K_{total} = $\frac{7}{10}$ 6.40 16.0² 0.130²

K_{total} = 19.4 J

6 0

2 years ago

How does the electric force between two charged particles change if one particle's charge is increased by a factor of 2?

Burka [1]

if one of the charge is doubled, the electric potential energy would be doubled too

8 0

3 years ago

Read 2 more answers