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Tpy6a [65]
3 years ago
10

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unch

anged?
Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

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When you hold a frozen banana treat in your hands, which statement BEST describes what is taking place? A. Heat is flowing from
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Answer:

The answer is A

Explanation:

It is A because your body heat is warmer than the banana and when you hold it the heat is transferring over.

3 0
3 years ago
Why are the parts of an atom that electrons occupy called electron clouds?
Anna [14]

Because it's literally impossible to tell exactly where something that size is
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And that's NOT because it's so small that we can't see it.  It's because any
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the more the size of its waves get to be like the same size as the object. 
When you get down to things the size of subatomic particles, it doesn't make
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cloud of <em><u>probability</u></em> of where the particle is.

I know it sounds weird.  But that's the way it is.  Sorry.


8 0
3 years ago
What are the 4 categories that makes the human body
Dominik [7]

Answer:

epithelial, muscle, nervous, and connective tissues.

Explanation:

3 0
3 years ago
A substance freezes at -58c therefore substance melts at ?
Yanka [14]
The freezing point is the same as the melting point.

If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
3 0
2 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
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