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Drupady [299]
3 years ago
8

Which elements do not usually interact with other elements?

Physics
2 answers:
d1i1m1o1n [39]3 years ago
7 0
Noble gasses ( insert gases)
kobusy [5.1K]3 years ago
6 0

Noble gases are the elements which do not usually interact with other elements.

Explanation:

The elements present in the last column of the periodic table defined as noble gases are the elements which do not interact with other elements.

The elements present in the noble gas series have fully occupied valence orbital and thus they are highly stable and does not undergo any kind of chemical reaction with other elements.

They are also termed as inert gases.

They are mostly used as ionizing gases as they can ionize the other gases easily.

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The skater is has a mass of 75 kg. Find the total Potential energy of the skater at the top of the ramp at 6 m.
AlekseyPX
P= mgh
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4 0
3 years ago
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
RSB [31]

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

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3 years ago
If asked to describe a wave, you could say waves are the
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Good evening Carolina


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3 0
3 years ago
Read 2 more answers
The strength of the gravitational force between two objects depends on________?
Gala2k [10]
Masses of the objects

3 0
3 years ago
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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
4 years ago
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