In fission reactions, how must the binding energy per nucleon vary? a. The binding energy per nucleon remains constant as atomic
2 answers:
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Answer:
Part A: The voltage across resistor R1 is approximately .
Part B: When the value of resistor R1 decreases, the current in this circuit will increase.
Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.
Explanation:
<h3>Part A</h3>Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of . The voltage across the two resistor, combined, is equal to
. Hence by Ohm's Law, the current through the circuit will be equal to
.
These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to . Apply Ohm's Law (again) to find the voltage across R1:
.
Since the equivalent resistance is equal to , when the value of
decreases, the equivalent resistance will also decrease. By Ohm's Law,
. When the value of the denominator ( decreases, the value of the quotient,
the current through the circuit, will increase.
Keep in mind that if two resistors are connected in series,
.
The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)
Again, since the two resistors are connected in series,
,
when the voltage across R2 increases, the voltage across R1 will decrease.
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.