Which components of an atom have a negative charge?

When very electronegative atoms, like oxygen, bond to atoms with lower electronegativity, like lithium, what's the result?

umka2103 [35]

(A) The oxygen atom becomes strongly negative.

Explanation:

When a very electronegative atom like oxygen bonds to an atom with a lower electronegativity, like lithium, the oxygen atom becomes strongly negative and the lithium becomes strongly positive.

The attraction between these oppositely charged ions results in the formation ionic bond.

An ionic bond is an interatomic bond in which a highly electronegative atom is bonded to less electronegative one.
The electronegative nature of the oxgyen allows is to gain the electrons lost by weakly electronegative lthium.
Most metals have low electronegativity.
They then both will form ions.
The attraction between the oppositely charged ions results in an ionic/electrovalent bonding.

learn more:

Ionic bond brainly.com/question/6071838

#learnwithBrainly

3 0

3 years ago

If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel

expeople1 [14]

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

$V = u + at$

Substituting into the equation, we have;

$V = 0 + 9.8 * 3.6$

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting the values into the equation;

$S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}$

$S = 0 + 4.9*12.86$

$S = 0.5 *36$

S = 63.50m

Therefore, the height of the tower is 63.50m.

6 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

The magnetic field around a magnet is called<br> Plss tell me fasssttt

Olenka [21]

Explanation:

The magnetic needle of a compass lines up with Earth's magnetic poles.

6 0

2 years ago

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=

True [87]

Answer:

(a) average velocity = 17.6 m/s

(b) when t = 0, v = 0

when t = 4, v = 19.2 m/s

when t = 8, v = 28.8 m/s

(c) after starting from rest, the car will be at rest again in 20 s

Explanation:

Given;

x(t)=bt²−ct³, substitute the given values and the equation will become;

x(t)=3t²−0.1t³

(a)average velocity = total distance / total time

total distance, x(t) = 3t²−0.1t³

x(8) = 3t²−0.1t³

X(8) = 3(8)² - 0.1(8)³

X(8) = 140.8 m

total time = 8 s

average velocity = 140.8 / 8

average velocity = 17.6 m/s

(b) instantaneous velocity = dx / dt

dx / dt = 6t - 0.3t²

when t = 0

v = 0

when t = 4 s

v = 6(4) - 0.3(4²) = 19.2 m/s

when t = 8 s

v = 6(8) - 0.3(8²) = 28.8 m/s

(c) the velocity is zero at dx / dt = 0

6t - 0.3t² = 0

t(6 - 0.3t) = 0

t = 0 or 6 - 0.3t = 0

t = 0 or 0.3t = 6

t = 0 or t = 6 / 0.3

t= 0 or t = 20 s

After starting from rest, the car will be at rest again in 20 s

6 0

3 years ago