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Art [367]
3 years ago
9

The greater the difference in electronegativity between two covalently bonded atoms

Physics
1 answer:
katrin [286]3 years ago
7 0

Answer:

The greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

Explanation:

Bond polarity (i.e the separation of electric charge along a bond) and ionic character (amount of electron sharing) increase with an increasing difference in electronegativity.

Therefore, we can say that, the greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

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You travel in a car 30 miles north to ava and home again. what is your distance? __________________________________________ what
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If you travel 30 miles somewhere and then come home again your distance is 60 miles.  Your displacement is 0 because it is the ending position minus the beginning position, which are the same place (home).  In other words, displacement is a vector and distance is a scalar.
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3 years ago
Filtration is used to separate mixtures based on _____.
sergejj [24]
Based on particle size.
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8 0
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Hii! help asap. i’ll give brainliest thanks!
o-na [289]

I believe it’s the mass of the box but I don’t no if I’m right

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5 0
3 years ago
A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end
Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0
3 years ago
A 2 kg, frictionless block is attached to a horizontal, ideal spring with spring constant 300 N/m. At t = 0 the spring is neithe
schepotkina [342]

Answer:

Explanation:

Given that,

Mass of block

M = 2kg

Spring constant k = 300N/m

Velocity v = 12m/s

At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0

xo = 0

It velocity is 12m/s at t=0

Then, it initial velocity is

Vo = 12m/s

Then, amplitude is given as

A = √[xo + (Vo²/ω²)]

Where

xo is the initial amplitude =0

Vo is the initial velocity =12m/s

ω is the angular frequency and it can be determine using

ω = √(k/m)

Where

k is spring constant = 300N/m

m is the mass of object = 2kg

Then,

ω = √300/2 = √150

ω = 12.25 rad/s²

Then,

A = √[xo + (Vo²/ω²)]

A = √[0 + (12²/12.5²)]

A = √[0 + 0.96]

A = √0.96

A = 0.98m

4 0
3 years ago
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