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3 years ago

If the atoms that share electrons have an unequal attraction for the electrons is called

1 answer:

6 0

If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.

<h3><u>Explanation:</u></h3>

A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.

HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : $H^{+} - Cl^{-}$ ). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.

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How long does it take an object to travel 375 m at a rate of 25 m/s?

Leni [432]

Answer:

15s

Explanation:

Given parameters:

Distance traveled = 375m

Speed = 25m/s

Unknown:

Time taken = ?

Solution:

To solve this problem, we make time the subject of the speed equation.

Speed = $\frac{distance}{time}$

Time = $\frac{distance}{speed}$

Now insert the parameters and solve;

Time = $\frac{375}{25}$ = 15s

3 0

3 years ago

3) If an object has a net negative charge of 4.0 Coulombs, the object possesses

Pani-rosa [81]

The answer to this question is option 2

3 0

3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

6 0

3 years ago

The image illustrates that as the distance between two objects increases, the force of gravity ____________. A) decreases. B) in

emmainna [20.7K]

The image is missing (however it's not necessary to solve the problem).

The correct answer is A) decreases, because the gravitational force is inversely proportional to the square of the distance. In fact, the magnitude of the gravitational force between two object of mass M and m, at a distance d one from each other, is
$F=G \frac{Mm}{d^2}$
where G is the gravitational constant. As can be seen from the formula, if the distance d between the two object increases, the intensity of the force decreases.

7 0

3 years ago

Read 2 more answers

10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave

zhannawk [14.2K]

Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,

f₁ = v/2L

where,

v = speed of wave = 36 m/s

L = Length of rope = 12 m

f₁ = fundamental frequency

Therefore,

f₁ = (36 m/s)/2(12 m)

f₁ = 1.5 Hz

Now the frequency of nth harmonic is given in general, as:

fn = nf₁

where,

fn = frequency of nth harmonic

n = No. of Harmonic = 2

f₁ = fundamental frequency = 1.5 Hz

Therefore,

f₂ = (2)(1.5 Hz)

5 0

3 years ago