An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:
Find the radius and period of the orbit.
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Answer:
r = 2,026 10⁹ m and T = 2.027 10⁴ s
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F =
Acceleration is centripetal
a = v² / r
we substitute
r = (1)
let's look for the charge in the insulating sphere
ρ = q₂ / V
q₂ = ρ V
the volume of the sphere is
v = 4/3 π r³
we substitute
q₂ = ρ π r³
q₂ = 3 10⁻⁹ π 4³
q₂ = 8.04 10⁻⁷ C
let's calculate the radius with equation 1
r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)
r = 2,026 10⁹ m
this is the radius of the electron orbit around the charged sphere.
Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio
v = x / t
the distance traveled in a circle is
x = 2π r
In this case, time is the period
v = 2π r /T
T = 2π r /v
let's calculate
T = 2π 2,026 10⁹/628 103
T = 2.027 10⁴ s
Related Questions
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Answer:
Earths atmosphere causes most small to medium meteors to burn up disintergrate before hitting the earth the moon does not have a protective atmosphere
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Answer:
-9Q
Explanation:
Electric field at origin is:
Electric field due to first charge at origin would be:
Electric field due to second charge would be:
If the second charge is Q', then should be:
compare the above two values to find the possible values of Q':
The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.
Thus, Q' = -9Q
One value is possible as the location of the second charge is given to be on the positive x-axis.
Final answer:
The possible values for the unknown charge are 1/9 of the magnitude of the known charge.
Explanation:
To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:
2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2
By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:
2 = 1 - 1/9
1/9 = 1
After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.
Learn more about Electric field here:
#SPJ3
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Answer:
67.5 cm
Explanation:
u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm
let f be the focal length
Use lens equation
1 / f = 1 / v - 1 / u
1 / f = 1 / 270 + 1 / 90
1 / f = 4 / 270
f = 67.5 cm
Final answer:
To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.
Explanation:
To determine the focal length of the lens, we can use the lens formula:
1/f = 1/v - 1/u
Where f is the focal length, v is the image distance, and u is the object distance.
Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:
1/f = 1/(3v) - 1/(90)
Multiplying through by 90*3v, we get:
90*3v/f = 270v - 90*3v
90*3v/f = 270v - 270v
90*3v/f = 0
Simplifying further, we find that: v = 0
When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.
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Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
Definition of wavelength
First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
Definition of frequency
On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Definition of propagation speed
Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.
Frequency of light waves with wavelength of 5×10⁻⁷ m
In this case, you know:
- v= 3×10⁸ m/s
- f= ?
- λ= 5×10⁻⁷ m
Replacing in the definition of propagation speed:
3×10⁸ m/s = f× 5×10⁻⁷ m
Solving:
3×10⁸ m/s ÷ 5×10⁻⁷ m= f
f= 6×10¹⁴ Hz
In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
Learn more about wavelength, frecuency and propagation speed:
brainly.com/question/2232652?referrer=searchResults
Answer:
Speed of light =m/s
wavelength = m
frequency = ?
we have
Speed = frequency × wavelength
= frequency ×
Frequency = hz
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Answer:
Statement 3 is correct.
Heisenberg's uncertainty principle explains that the measurement of an observable quantity in the quantum domain inherently changes the value of that quantity
Explanation:
Classical mechanics is the study of motion of big, relatable bodies that we come in contact with in our day to day lives.
Quantum mechanics refers to this same study, but for particles on a subatomic level.
Obviously, Classical mechanics' theories and principles were first discovered and they worked for their intended uses (still work!). But when studies on particles on a sub-atomic level intensified, it became impractical to apply those theories and principles to these sub-atomic particles that displayed wave-particle duality nature properly.
Heisenberg's Uncertainty principle came in a time that explanations and justifications were needed to adapt these theories to sub-atomic particles.
The principle explains properly that it is impossible to measure the position and velocity (momentum) of a sub-atomic particle in exact terms and at the same time.
Mathematically, it is presented as
Δx.Δp ≥ ℏ
Where ℏ= adjusted Planck's constant.
ℏ= (h/2π)
And Δx and Δp are the uncertainties in measuring the position and momentum of sub-atomic particles.
The major reason for this is the wave-particle duality of sub-atomic particles. They exist as waves and particles at the same time that a complete knowledge of their position mean that a complete ignorance of their velocity and vice versa.
Taking the statements one at a time
Statement 1
Quantum Mechanics studies sub-atomic particles which are mostly always in motion. So, this is false.
Statement 2
It is impossible to calculate with accuracy both the position and momentum of particles in quantum mechanics not classical mechanics. As stated above, the reason for the uncertainty is the wave-particle duality of sub-atomic particles which the particle in classical mechanics do not exhibit obviously enough.
Statement 3
Any attempt to measure precisely the velocity of a subatomic particle, will knock it about in an unpredictable way, so that a simultaneous measurement of its position has no validity.
An essential feature of quantum mechanics is that it is generally impossible, even in principle, to measure a system without disturbing it. This is basically the uncertainty principle rephrased. This is the only true statement.
Hope this Helps!!!