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OlgaM077 [116]
3 years ago
14

What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An

swer in units of eV.
Physics
1 answer:
GREYUIT [131]3 years ago
4 0

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}

E=\frac{2m_{0}c^}2}{\sqrt{3}}

E=\frac{2\times0.001\times9\times10^{16}}{1.732}

E=1.04\times10^{14}J

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV

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An elevator of massmis initially at rest onthe first floor of a building. It moves upward,and passes the second and third floors
Varvara68 [4.7K]

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

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Where,

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When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

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Replacing in our equation we have to

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The correct answer is 4.

7 0
3 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

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