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snow_tiger [21]
3 years ago
10

Which statement is true? Which statement is true? An orbital that penetrates into the region occupied by core electrons is more

shielded from nuclear charge than an orbital that does not penetrate and therefore has a higher energy. An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a higher energy. An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy. An orbital that penetrates into the region occupied by core electrons is more shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy.
Physics
1 answer:
masya89 [10]3 years ago
6 0

Hey there!:

Here the Statement - D is correct.  

Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate.  Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.

Hope this helps!

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How is lightning formed with explanation l?
creativ13 [48]
Lightning is an electric current. Within a thundercloud way up in the sky, many small bits of ice (frozen raindrops) bump into each other as they move around in the air. All of those collisions create an electric charge. After a while, the whole cloud fills up with electrical charger
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What are main causes by tornadoes
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Updrafts and downdrafts in a thunderstorm interacting with wind shear
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3 years ago
A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient
e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

\displaystyle \mu=\frac{80}{196}

\mu=0.41

The coefficient of static friction between the crate and the floor is 0.41

7 0
3 years ago
A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
3 years ago
A student tests the acidity of nitric acid (HNO3) by dissolving a sample of HNO3 in a solution of liquid methane. He then uses a
DerKrebs [107]
Sorry I didn't see this before...
Okay, I see two major problems with this student's experiment:

1) Nitric acid Won't Dissolve in Methane
Nitric acid is what's called a mineral acid. That means it is inorganic (it doesn't contain carbon) and dissolves in water.
Methane is an organic molecule (it contains carbon). It literally cannot dissolve nitric acid. Here's why:
For nitric acid (HNO3) to dissolve into a solvent, that solvent must be polar. It must have a charge to pull the positively charged Hydrogen off of the Oxygen. Methane has no charge, since its carbon and hydrogens have nearly perfect covalent bonds. Thus it cannot dissolve nitric acid. There will be no solution. That leads to the next problem:

2) He's Not actually Measuring a Solution
He's picking up the pH of the pure nitric acid. Since it didn't dissolve, what's left isn't a solution—it's like mixing oil and water. He has groups of methane and groups of nitric acid. Since methane is perfectly neutral (neither acid nor base), the electronic instrument is only picking up the extremely acidic nitric acid. There's no point to what he's doing.

Does that help?
8 0
3 years ago
Read 2 more answers
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