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3 years ago

An electron is moving in a circular orbit in a uniform magnetic field. Is the kinetic energy of the electron changing?

Physics

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

kinetic energy does not change

Explanation:

you can use the formula for the kinetic energy of the electron and for the radius of the trajectory of an electron in a uniform magnetic field:

$E_k=\frac{1}{2}m_ev^2\\\\r=\frac{m_e v}{qB}$

me: mass of the electron

B: magnetic field

q: charge of the electron

r: radius

By doing v the subject of the formula and replace it in the expression for the kinetic energy you obtain:

$v=\frac{rqB}{m_e}\\\\E_k=\frac{1}{2}m_e(\frac{rqB}{m_e})^2=\frac{r^2q^2B^2}{2m_e}$

as youcan see, all parameters r, q, B and me are constant.

hence, the kinetic energy does not change

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Answer:

B. chemical only.

Explanation:

In the process of respiration which is a chemical process where organic compound is released. In this process exergonic reaction takes place in which compound changes into different ones.

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3 years ago

A speaker vibrates at a frequency of 200 hz what is its period

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3 years ago

Read 2 more answers

The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7

stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

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3 years ago

The mass of a star is 1.61·1031 kg and its angular velocity is 1.60E-7 rad/s. Find its new angular velocity if the diameter sudd

Tcecarenko [31]

Answer:

ω₂ = 1.9025 x 10⁻⁶ rad/s

Explanation:

given,

mass of star = 1.61 x 10³¹ kg

angular velocity = 1.60 x 10⁻⁷ rad/s

diameter suddenly shrinks = 0.29 x present size

r₂ = 0.29 r₁

using conservation of angular momentum

I₁ ω₁ = I₂ ω₂

$(\dfrac{2}{5}mr_1^2)\omega_1=(\dfrac{2}{5}mr2^2)\omega_2$

$r_1^2\times \omega_1=r_2^2\times \omega_2$

$r_1^2\times 1.60\times 10^{-7}=(0.29r_1)^2\times \omega_2$

$1.60\times 10^{-7}=0.0841\times \omega_2$

$\omega_2=\dfrac{1.60\times 10^{-7}}{0.0841}$

ω₂ = 1.9025 x 10⁻⁶ rad/s

5 0

3 years ago