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3 years ago

How do magnetic forces repel or attract?

1 answer:

3 0

“Magnets are surrounded by an invisible magnetic field that is made by the movement of electrons, the subatomic particles that circle the nucleus of an atom”

“Every magnet has both a north and a south pole. When you place the north pole of one magnet near the south pole of another magnet, they are attracted to one another. When you place like poles of two magnets near each other (north to north or south to south), they will repel each other.”

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Find the de Broglie wavelength of the following. (a) a 4-MeV proton 14.3 Correct: Your answer is correct. fm (b) a 40-GeV electr

kherson [118]

a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

The de Broglie wavelength of an object is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the particle

Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

$m_0 = 938 MeV$

And since $4MeV$ , this means that the proton is non-relativistic. So its kinetic energy is related to its momentum by

$E=\frac{p^2}{2m}$

which means

$p=\sqrt{2Em}$

where

$E=4 MeV \cdot 10^6 eV/MeV \cdot 1.6\cdot 10^{-19] J/eV=6.4\cdot 10^{-13} J$ is the kinetic energy

$m=1.67\cdot 10^{-27} kg$ is the proton mass

Substituting, we find

$\lambda=\frac{h}{\sqrt{2Em}}=\frac{6.63\cdot 10^{-34}}{\sqrt{2(6.4\cdot 10^{-13})(1.67\cdot 10^{-27})}}=14.3\cdot 10^{-15} m = 14.3 fm$

In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

$m_0 = 0.511 MeV$

Since $40 GeV >> 0.511 MeV$ , the electron is ultra-relativistic: so we can rewrite its energy as

$E = pc$

The equation (1) can also be rewritten as

$\lambda = \frac{hc}{pc}$

where c is the speed of light. The quantity at the denominator is the energy, so

$\lambda=\frac{hc}{E}$

where:

$E=40 GeV = 40\cdot 10^9 eV \cdot (1.6\cdot 10^{-19})=6.4\cdot 10^{-9} J$ is the energy of the electron

And substituting, we find:

$\lambda=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{6.4\cdot 10^{-9}}=3.1 \cdot 10^{-17} m = 0.031 fm$

Learn more about de Broglie wavelength:

brainly.com/question/7047430

#LearnwithBrainly

4 0

3 years ago

The bulge of water on the side of earth closest to the moon produces _______.

Goryan [66]

That's a "high tide".

6 0

3 years ago

In first equation of motion at denotes

RideAnS [48]

Answer:

Names of the Equations of Motion

1 The First Equation of Motion is, v=u+at is known as velocity – time relation. 2 The Second Equation of Motion is, s=ut+12at2 is known as position – time relation.

6 0

3 years ago

What is static Friction?

postnew [5]

Exists between a stationary object and the surface on which it's resting

5 0

4 years ago

Read 2 more answers

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

Minchanka [31]

The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression.
In general, the potential energy of gravitational field is defined as:
$U=-G \frac{mM}{r}$
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
$F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}$
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
$g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg$
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
$U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}$
When we plug in the numbers we get:
$U=-4.99\cdot 10^{10} J$
The potential energy has to be equal to the kinetic energy.
$E_k=4.99\cdot 10^{10} J$

3 0

3 years ago