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SSSSS [86.1K]
2 years ago
5

Which nucleus completes the following equation?

Physics
2 answers:
IgorLugansk [536]2 years ago
4 0

Answer: ^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr

The chromium nucleus will complete the reaction.

Explanation:

The given reaction is a type of radioactive nuclei decay:

Positron emission: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as \beta ^+-decay. In this the mass number remains same.

_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e

Now according to the given reaction:

^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr

The chromium nucleus will complete the reaction.

Olegator [25]2 years ago
4 0

Answer:

50 25 Mn-0 1 e+ 50 24 Cr

Explanation:

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Answer:

D an B

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A runner has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?
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I could be wrong, but I'm pretty sure it's 144kg.
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An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco
sdas [7]
A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
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Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
8 0
2 years ago
A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc
Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is 6.21 \times 10^{-3} N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = 6.50 \times 10^{-6} C

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    F_{net} = 6.25 \times 10^{-3} N in the +x direction

Also,       F_{net} = F_{E} - F_{b}

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = (\frac{1}{B}) \times (E - \frac{F_{net}}{q})

                 = (\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for F_{net}.

Using the right hand rule where:

your right thumb goes toward the F_{net}, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that v_{x}, v_{y}, v_{z} = 0, -(458.507), 0.

8 0
2 years ago
A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field
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magnetic field due to a finite straight conductor is given by

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here since it forms an equilateral triangle so we will have

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also the perpendicular distance of the point from the wire is

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now from the above equation magnetic field due to one wire is given by

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B = \frac{3\mu_0 i}{2\pi a}

now since in equilateral triangle there are three such wires so net magnetic field will be

B = \frac{9\mu_0 i}{2\pi a}

5 0
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