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Lelu [443]
3 years ago
9

Am I the only one or do I hate scooping the cat box if you guys have cats.​

Physics
2 answers:
LenaWriter [7]3 years ago
4 0
THIS DUDE REALLY SHOWED US CAT TURDS...... IM DYING LM.AOOOOOOOO
BAHAHAHA
nordsb [41]3 years ago
4 0

Answer:

yeah we have a self cleaning litter box

Explanation:

i kinda hate cats

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List three reasons that rain predictions made by meteorologists are sometimes incorrect. Explain your response.
Agata [3.3K]
1. <span>the low pressure is moving slower than expected.
This make the meteorologist receive premature data which make them fail to interpret the data correctly and make the wronf prediction.
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Describe the octet rule in terms of noble-gas configurations and potential energy
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Elements will gain or lose electrons to form a noble gas configuration. Atoms that meet the octet rule are stable because their valence electrons have a relatively low potential energy. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.


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3 years ago
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How are animals of the coniferous forest well adapted to long, cold winters?
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They have thick body coverings
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3 years ago
What are noble gases?
Bas_tet [7]

Answer:

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6 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
3 years ago
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