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2 years ago

Use the periodic table and your knowledge of isotopes to complete these statements.

2 answers:

7 0

Use the periodic table and your knowledge of isotopes to complete these statements.

When polonium-210 emits an alpha particle, the child isotope has an atomic mass of ⇒ 206.

I-131 undergoes beta-minus decay. The chemical symbol for the new element is ⇒ Xe.

Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of ⇒ 18.

sleet_krkn [62]2 years ago

7 0

Answer:

206,Xe,18

Explanation:

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A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 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

Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t

Doss [256]

Answer:

The charge to mass ratio is $-1.76\times10^{11}$

Explanation:

$Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\$

We need to find how much charge is contained in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

$\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}$

4 0

2 years ago

Read 2 more answers

Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This r

alexandr1967 [171]

Answer:

to create the particle the speed must be greater than 2.25 10⁸ m / s

Explanation:

In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium

n = c / v

where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio

in this case they give us that the medium matter water them that has a refractive index of

n = 1,333

we clear

v = c / n

let's calculate

v = 3 10⁸ / 1,333

v = 2.25 10⁸ m / s

to create the particle the speed must be greater than 2.25 10⁸ m / s

6 0

2 years ago

The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to

Taya2010 [7]

Answer:

B. i=2.79A

C. F=0.066N

Explanation:

A) By the right hand rule we have that

F=iL x B

F=iLBsin(α)

If the wire jump toward the observer the top pole face is the magnetic southpole.

B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

$F=iLBsin(\alpha)\\\alpha=90°\\F=iLB\\i=\frac{F}{LB}=\frac{6.71*10^{-2}N}{(0.15m)(0.16T)}=2.79A$

C) Now the length of the wire that feels B is

$L=\frac{0.15m}{cos(10\°)}=0.152m$

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

$F=iLBsin(\alpha)\\F=(2.79A)(0.152m)(0.16T)(sin(80\°))=0.066N$

I hope this is useful for you

regards

8 0

2 years ago