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2 years ago

How do the nuclei of covalently bonded atoms help keep the bond together?

Physics

2 answers:

Mkey [24]2 years ago

5 0

Answer:

the answer is d

Explanation:

and BOOM! the right answer!

kifflom [539]2 years ago

3 0

Answer;

D. Positive particles in the nucleus are attracted to shared electrons, so the atoms stay close together

Explanation;

-Covalent bonds hold atoms together because the attraction between the positively charged nuclei and the negatively charged shared electrons is greater than the repulsions between the nuclei themselves.

-The attraction between the positively charged nuclei and the negatively charged shared electrons is greater than the repulsions between the nuclei themselves.

-As two atoms approach each other, the electrons in their outer shells start to notice the nucleus of the other atom. As a result, they become attracted not only to their own nucleus, but to the nucleus of the other atom as well. The electrons then tend to spend more of their time in between the two nuclei, instead of just around their own nucleus.

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If the frequency of a given wave increases,what happens to the wavelength?

Greeley [361]

It shortens so that the tips reach faster

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2 years ago

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Is a perch a consumer or...

labwork [276]

Yes a perch is a consumer.

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3 years ago

A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration

Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

$F_f = \mu N$

Where,

$\mu =$ Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

$m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\$

By condition of Balance the friction force must be equal to the total net force, that is to say

$F_{net} = F_f$

$m_{total}a = \mu m_hg$

$(m_h+m_c)a = \mu*m_h*g$

Re-arrange to find acceleration,

$a= \frac{\mu*m_h*g}{(m_h+m_c)}$

$a = \frac{0.894*242*9.8}{(242+224)}$

$a = 4.54 m/s^2$

Therefore the acceleration the horse can give is $4.54m/s^2$

3 0

2 years ago

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f

Usimov [2.4K]

Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

$\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s$

Therefore, the rotational speed of the ring module have to be 0.92 rad/s

3 0

3 years ago

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

Semenov [28]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

3 0

2 years ago