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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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Explanation:

\law \: of \: conservation \: of \: momentum)

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3 years ago
A wrench is placed at 30 cm in front of a diverging lens with a focal length of magnitude 10 cm. What is the magnification of th
WARRIOR [948]

Answer:

0.25

Explanation:

Magnification = image distance/object distance

mag = v/u.................. Equation 1

Given: f = -10 cm ( diverging lens) u = 30 cm.

Where can calculate for the value of v using

1/f = 1/u+1/v

make v the subject of the equation

v = fv/(u-f)..................... Equation 2

Substitute into equation 2

v = -30(10)/(30+10)

v = -300/40

v = -7.5 cm.

substituting into equation 1,

mag = 7.5/30

mag = 0.25

hence the magnification of the wretch = 0.25

4 0
3 years ago
1 point
Darya [45]

Given that,

Mass of object, m = 55 kg

Mechanical energy of the object, M = 4306 J    

Potential energy, P = 2940 J

We know that the mechanical energy is the sum of kinetic and potential energy such that,

Mechanical energy = kinetic energy + potential energy

K=M-P\\\\K=4306-2940\\\\K=1366\ J

Kinetic energy is given by :

K=\dfrac{1}{2}mv^2v is velocity of object

v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1366}{55}} \\\\v=7.04\ m/s

So, the velocity of object is 7.04 m/s.            

7 0
3 years ago
A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it
Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}

\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}

\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

8 0
3 years ago
In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead
fiasKO [112]

Answer:

8.01e-22

Explanation:

7 0
3 years ago
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