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vivado [14]
3 years ago
11

Suppose we have three hypothetical atomic nuclei: A, B, and C. Nucleus A has 7 protons and 7 neutrons. Nucleus B has 5 protons a

nd 7 neutrons. Nucleus C has 7 protons and 5 neutrons. Which of the two nuclei are isotopes of the same element
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

Nucleus A with 7 protons and 7 neutrons and Nucleus C with 7 protons and 5 neutrons are isotopes of the same elements

Explanation:

Isotopes are elements that have the same atomic structure but different molecular structure. An atom that has the same atomic number but different mass number are known to be isotopes.

The proton of an atom is the same as its atomic number while the sum of the proton and neutron is equal to its mass number.

According to the question, nuclei that has the same number of proton are isotopes of the same element. Therefore nuclei A and C with 7 protons each are isotopes of the same element since they have the same atomic number i.e number of proton = atomic number.

Their atomic masses of nuclei A and C are 14 and 12 respectively

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The linear charge density on the inner conductor is and the linear charge density on the outer conductor is
Lubov Fominskaja [6]

Complete Question

The complete question is  shown on the first uploaded image (reference for Photobucket )

Answer:

The  electric field is  E = -1.3 *10^{-4} \ N/C

Explanation:

 From the question we are told that

    The linear charge density on the inner conductor is  \lambda _i  =  -26.8 nC/m  =  -26.8 *10^{-9} C/m

    The linear charge density on the outer conductor is

  \lambda_o  = -60.0 nC/m  =  -60.0 *10^{-9} \ C/m

     The position of interest is r =  37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field  at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

         E (2 \pi r l) =  \frac{ \lambda_i }{\epsilon_o}

=>    E =  \frac{\lambda _i }{2 \pi *  \epsilon_o * r}

substituting values  

       E =  \frac{ -26 *10^{-9} }{2 * 3.142 *  8.85 *10^{-12} * 0.0373}

       E = -1.3 *10^{-4} \ N/C

The  negative  sign tell us that the direction of the electric field is radially inwards

    =>   |E| = 1.3 *10^{-4} \ N/C

5 0
4 years ago
The hydraulic crane is used to lift the 1400-lb load. determine the force in the hydraulic cylinder ab and the force in links ac
zlopas [31]
<span>Answer: So at joints C and D do I let there be a Cx, Cy and Dx, Dy respectively or do I just keep it as Fwd and Fwc? Either way I can't seem to solve it. â‘Fx=0=WD-4293cos19.74 ÎŁFy=0=Fwc Sin19.74 - 1450 Fwc= 4203 lb WD = 4041 lb Then at joint C â‘Fx=0=CDsin30 - CAsin30 - 4041 â‘Fy=0=1450-CDcos30-CAcos30 CD = 4878 CA = -3204 Incorrect. Reference https://www.physicsforums.com/threads/statics-hydraulic-crane.776033/</span>
6 0
4 years ago
Why is important the electromagnetism, diff usages, and how will be our World without it
lawyer [7]

Answer:

Electromagnetism has important scientific and technological applications. It is used in many electrical appliances to generate desired magnetic fields. It is even used in a electric generator to produce magnetic fields for electromagnetic induction to occur. It has many more technological applications including MRI scanning (magnetic resonance imaging) and electric bells.

Explanation:

Electromagnets are widely used as components of other electrical devices, such as motors, generators, electromechanical solenoids, relays, loudspeakers, hard disks, MRI machines, scientific instruments, and magnetic separation equipment.

7 0
3 years ago
Why is Alternating Current important
otez555 [7]
It’s voltage can easily be modified and it allows power to be transmitted at a high voltage rate before being lowered to a smaller voltage rate to be safe.
6 0
3 years ago
Read 2 more answers
A very long wire carries a uniform linear charge density of 7.0 nC/m. What is the electric field strength 16.0 m from the
jok3333 [9.3K]

Electric field due to a long wire is given by

E = \frac{2 k \lambda}{r}

here

k =  9 * 10^9

\lambda = 7*10^{-9}

r = 16 m

E = \frac{2*9*10^9 *7*10^{-9}}{16}

E = 7.875 N/C

6 0
4 years ago
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