I think the correct answer is the second option. A circuit describes a closed conducting loop through which an electrical current can flow. It is a path that an electrical current could flow. A circuit could be a closed one or an open circuit. A closed circuit would be a circuit where the current could flow continuously. An open circuit would be a type of circuit where the flow current would only go once and stopped at a particular point since the current has nowhere to go. For a circuit to work, an electric supply should be available to supply the electric current.
Answer:
0.22 b
Explanation:
Quadrupole moment of the nucleon is,
![Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}](https://tex.z-dn.net/?f=Q%3D-%5Cfrac%7B2j-1%7D%7B2%28j%2B1%29%7D%5Cfrac%7B3%7D%7B5%7DR%5E%7B2%7D)
And also,
![R^{2}=R^{2} _{0}A^{\frac{2}{3} }](https://tex.z-dn.net/?f=R%5E%7B2%7D%3DR%5E%7B2%7D%20_%7B0%7DA%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%7D)
And, ![R _{0}=1.2\times 10^{-15}m](https://tex.z-dn.net/?f=R%20_%7B0%7D%3D1.2%5Ctimes%2010%5E%7B-15%7Dm)
Now,
![Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }](https://tex.z-dn.net/?f=Q%3D-%5Cfrac%7B2j-1%7D%7B2%28j%2B1%29%7D%5Cfrac%7B3%7D%7B5%7DR%5E%7B2%7D%20_%7B0%7DA%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%7D)
For Bismuth
and A is 209.
![Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn](https://tex.z-dn.net/?f=Q%3D-%5Cfrac%7B2%5Cfrac%7B9%7D%7B2%7D%20-1%7D%7B2%28%5Cfrac%7B9%7D%7B2%7D%20%2B1%29%7D%5Cfrac%7B3%7D%7B5%7D%281.2%5Ctimes%2010%5E%7B-15%7D%29%20%5E%7B2%7D%28209%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%20%7D%5C%5CQ%3D0.628%5Ctimes%2035.28%5Ctimes%2010%5E%7B-30%7D%20%5C%5CQ%3D22.15%5Ctimes%2010%5E%7B-30%7D%20m%5E%7B2%7D%20%5C%5CQ%3D0.2215%5Ctimes%2010%5E%7B-28%7D%20m%5E%7B2%7D%20%5C%5CQ%3D0.22%20barn)
Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b
<span>Data:
mass =
110-g bullet
d = 0.636 m
Force =
13500 + 11000x - 25750x^2, newtons.
a) Work, W
W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =
W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =
W = 8602.6 joule
b) x= 1.02 m
</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02
W = 10383.5
c) %
[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.
</span>
Answer:
c)the gravitational forces of people is so small it is overshadowed by that of earth.
Explanation:
The gravitational force between two objects is given by:
![F=G\frac{m_1 m_2}{r^2}](https://tex.z-dn.net/?f=F%3DG%5Cfrac%7Bm_1%20m_2%7D%7Br%5E2%7D)
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the distance between the two objects
From the formula, we see that the gravitational force depends on the masses of the objects: since the mass of the Earth (
is much much larger than the average mass of one person (80-100 kg), then the gravitational force exerted by the Earth on a person is also much much larger than the gravitational force between two people.
The Equivalent resistance is :
![\qquad \tt \dashrightarrow \: \dfrac{14}{9} \: \: ohms](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctt%20%5Cdashrightarrow%20%5C%3A%20%5Cdfrac%7B14%7D%7B9%7D%20%5C%3A%20%20%5C%3A%20ohms)
The solution is in attachment for solution ~