1) The motor is 75% efficient.
1 answer:
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Answer:
The wavelength of the light decreases as it enters into the medium with the greater optical density.
The frequency of the light remains constant as it transitions between materials.
Explanation:
- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:
where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.
- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement
The frequency of the light remains constant as it transitions between materials.
is correct.
- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation
where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement
The wavelength of the light decreases as it enters into the medium with the greater optical density.
is also correct.
Answer:
a. It starts at point B.
vp = 2.53*10⁴ m/s
a. it starts at point A.
ve= 1.08*10⁶ m/s
Explanation:
a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.
Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:
ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)
⇒
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.
Replacing by these values, and solving for v, we have:
⇒ vp = 2.53*10⁴ m/s
b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:
First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.
Second, as its charge is (-e) the change in electric potential energy had been negative also:
ΔUe = -e*ΔV = -e* (VB-VA)
In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.
Replacing by these values, and solving for v, we have:
⇒ ve = 1.08*10⁶ m/s