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torisob [31]
3 years ago
5

An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a

re perpendicular to each other, what is the electron's speed
Physics
1 answer:
lesantik [10]3 years ago
8 0

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s

So, the speed of the electron is 2591.83 m/s.

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Consider a one-dimensional crystal (similar to a carbon nanowire) with length 10 um and lattice spacing 0.1 nm.
Finger [1]

Answer:

a)  Fermi level = 600 electron-volts

b) \frac{2.04 * 10^{13} }{\sqrt{E} }

Explanation:

Given data:

length of one-dimensional crystal = 10 um

Lattice spacing = 0.1 nm

A) Determine the Fermi level assuming one electron per atom

Total length = 10 <em>u</em>m

Interatomic separation of a = 0.1 nm

in this case the Atom has one electron therefore the number of electrons = 10^5  and the number of states Ns = gsN = 2 * 10^5  ( attached below is some part of the solution )

hence : Fermi level = 600 electron-volts

B) Determine the density of states as a function of electron energy

attached below is the detailed solution

5 0
3 years ago
its a hot day at the beach you look at your portable themometer which says its 30°C. what is the temperature in Fahrenheit and K
Nitella [24]

The answer is 86 degrees Fahrenheit.

Hope this helps :)

6 0
3 years ago
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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after
s2008m [1.1K]

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

6 0
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The nearest star to our planet other than the sun , is 4.4 light year away.One light-year is the distance light travels in a yea
lozanna [386]

Answer:

1606

The main reason for using light years, however, is because the distances we deal with in space are immense. If we stick to miles or kilometers we quickly run into unwieldy numbers just measuring the distance to the nearest star: a dim red dwarf called Proxima Centauri that sits a mere 24,000,000,000,000 miles away!

Explanation:

3 0
3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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