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Evgen [1.6K]
4 years ago
14

Why does the band of stability curve upward at high atomic numbers? excess neutrons are required due to the repulsion between th

e protons. atoms with high atomic numbers have a large number of electrons orbiting the nucleus. this increased number of electrons requires a lot of extra mass in the nucleus to keep the electrons in their orbit?
Physics
1 answer:
ella [17]4 years ago
3 0
The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.

Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.


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jeyben [28]
Balanced. They’re equally as strong so as their arm wrestling, neither of the men’s hands go down. Because they’re equally/balanced as strong.
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3 years ago
Drag the tiles to the correct boxes to complete the pairs.
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3 years ago
a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a
Whitepunk [10]

Answer:

W = -2.76\times 10^{-9}~J

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

U = \frac{1}{2}CV^2

where C can be calculated using the plate separation and area.

C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon

Therefore, the potential energy in the first case is

U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon

In the second case:

C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J

6 0
3 years ago
You wad up a piece of paper and throw it into the wastebasket. How far will
astraxan [27]

Answer:

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air resistance left we assume there isn't any, it might

just as well be a stone that's tossed.  This is just a

stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what.  I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is  1.4 sin(36)

                                                                        = 0.823 m/s up.

-- The projectile rises for (0.823/9.8) second, runs out of gas,

and then falls for another (0.823/9.8) second to its original height.

So it's in the air for

                                  2 (0.823/9.8) = 0.168 second

                                                            (not very long at all)

-- The horizontal component of the velocity is  1.4 cos(36)

                                                                           = 1.133 m/s  

                                                             and it doesn't change.

-- During the 0.168 second that it's in the air,

the wad travels horizontally

                                              (0.168 s) x (1.133 m/s)

                                          =            0.19 meter

                                              (19 cm, ~ 7.5 inches)

If you find my mistake on this one, please please tell me.  

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

6 0
3 years ago
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mafiozo [28]
Now really ! 

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