Whole grains are good carbs so it would be true.
And the easiest would be carbohydrates.
Answer:
Explanation:
Due to heat energy , metal expands . Formula for linear expansion is as follows .
L = l ( 1 + α Δt )
where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .
To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are
20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .
For steel sphere ,
L = 25 ( 1 + 12 x 10⁻⁶ Δt )
For brass ring
L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )
1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt
.004 = 7.9518 x 10⁻⁶ Δt
Δt = 4000 / 7.9518
= 503⁰C.
final temp = 503 + 15 = 518⁰C .
The scientific method. When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
Answer:
The correct answer is ![2.8*10^{-5}ms^{-1}](https://tex.z-dn.net/?f=2.8%2A10%5E%7B-5%7Dms%5E%7B-1%7D)
Explanation:
The formula for the electron drift speed is given as follows,
![u=I/nAq](https://tex.z-dn.net/?f=u%3DI%2FnAq)
where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.
Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is
. Converting this number to m³ using very elementary unit conversion we get
. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,
![n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}](https://tex.z-dn.net/?f=n%3D140384%2A6.02%2A10%5E%7B23%7D%20%3D%208.45%2A10%5E%7B28%7Delectrons.m%5E%7B-3%7D)
if we convert the area from mm³ to m³ we get
.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,
![u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}](https://tex.z-dn.net/?f=u%3D30%2F%288.45%2A10%5E%7B28%7D%2A80%2A10%5E%7B-6%7D%2A1.602%2A10%5E%7B-19%7D%29%5C%5Cu%3D2.8%2A10%5E%7B-5%7Dm.s%5E%7B-1%7D)
which is our final answer.
Proton positive; electron negative; neutron no charge<span>. </span>The charge<span> on the proton and </span>electron<span> are exactly the same size but opposite. The same number of protons and </span>electrons<span> exactly cancel one another in a neutral atom.
</span>
hoped it helped