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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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Which of the following objects is in static equilibrium?

DedPeter [7]

Answer:

b- a bicycle sitting on the ground

Explanation:

'Static' means it's not moving.

So, only the answer listed as "a bicycle sitting on the ground" corresponds to that condition.

All others possible answers describe something that is moving (an apple falling, a motorcycle accelerating, a car moving at a constant speed). Even if they are going at a constant speed, they are moving, so not static.

6 0

3 years ago

A 1750kg bumpercar moving at 1.50m/s to the right collides elastically with a 1450kg car going to the left at 1.10m/s. The 1750k

damaskus [11]

1984.08 kg that’s the answer

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3 years ago

A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy $U=4.30\times10^{3}\ J$

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

Where, $\Delta U$ = internal energy

T = average temperature

Put the value in to the formula

$\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}$

$\Delta S=-14.72\ J/K$

Hence, The approximate change in entropy is -14.72 J/K.

5 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

4 years ago

Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with

wlad13 [49]

It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.

4 0

3 years ago