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pshichka [43]
4 years ago
12

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

c to 0.942c MeV
Physics
1 answer:
Aleonysh [2.5K]4 years ago
4 0

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

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Which statement describes a property of a magnet? A. It attracts ferrous materials. B. It could have only one pole (north or sou
hodyreva [135]

Answer:

It attracts ferrous materials

Explanation:

A magnet attracts ferrous materials A ferrous materials are metallic substances or conductors that can conduct heat and electricity. Example of this ferrous materials includes iron, metal etc. Since magnets only can attracts metallic substance to itself, then we can also conclude that they attract ferrous materials since ferrous materials. possesses properties of a metal.

Magnets possesses both north and south poles.

The same of the bar magnets are known to repel each other while unlike poles attract each other.

7 0
3 years ago
An object of mass 20kg is released from a height of 10 meters above the ground level. The kinetic energy of the the object just
Yuri [45]

Answer:

The answer is the object weighs 5 so 3-5 is 2

Explanation:

I took this test

7 0
3 years ago
¿Cuál es la frecuencia de una ola con una velocidad de 14 m / s y una longitud de onda de 20 metros?
Margaret [11]

Responder:

<h2>0.7Hertz </h2>

Explicación:

Usando la fórmula para calcular la velocidad de onda que se expresa como se muestra.

Velocidad de una onda = frecuencia * longitud de onda

v = fλ

Dada la velocidad de onda = 14 m / sy longitud de onda = 20 metros

frecuencia f = v / λ

f = 14/20

f = 0.7Hertz

La frecuencia de la onda es de 0.7 Hertz.

7 0
3 years ago
When an electron moves 2.5 m in the direction of an electric field, the change in electrical potential energy of the electron is
anygoal [31]
In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,

E = V/d

However, the voltage in volts is energy per coulomb. Thus,
V = (8x10-17 J/electron)*(1electron/1.60218x10^-19 C)
V = 499.32 volts

Therefore,

E = 499.32 volts /2.5 m
E = 199.73 N/C

The electric field that caused the change in potential energy is equal to 199.73 Newtons per Coulomb.
7 0
3 years ago
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0
3 years ago
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