Based on the given question, the resistance will: increase with the increase in voltage.
The reason behind this is that resistance and voltage have a direct relationship. As the voltage increases, the resistance also increases. This can be explained by Ohm’s Law which states that V= IR where V is voltage, I is current and R is resistance. As per the second part of the question, it is implied that the resistance varies with temperature.
The resistance of any material depends upon temperature, and a rise in temperature increases the resistance of the material. The light bulb acts as a resistor, and its resistance will increase as the temperature increases due to an increase in the temperature of the filament of the bulb.
The resistance is directly proportional to the temperature of the bulb, and it is represented by the equation
R = R₀ (1 + αt),
where R is resistance, R₀ is the resistance at a particular temperature, α is the temperature coefficient of resistance, and t is the temperature difference in Celsius.
Therefore, based on the data provided, it can be concluded that resistance increases with the increase in temperature which results in the heating of the light bulb, which is a resistor.
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the surface of the sun appears sharp in visible light because
"The surface of the sun appears sharp in visible light because the photosphere is thin compared to the other layers in the sun."
Most of the electromagnetic energy that reaches the earth begins in the photosphere, the area of the sun that is visible to us. The photosphere is referred to as the sun's surface, despite the fact that it is a gaseous entity.
The gas in the photosphere appears to have a sharp surface, but in reality, it is heavier lower in the Sun and less dense higher up. It is more transparent the less thick it is. The area of the gas that is visible to us is where it has largely become translucent. About 300 km of this layer are deep.
The photosphere is the line separating the core of the Sun from its atmosphere. It is the part of the Sun's surface that is visible to us. The photosphere is not like a planet's surface; even if you could stand in the sun, you couldn't do so on the photosphere.
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in one cycle a heat engine absorbs 480 j from a high-temperature reservoir and expels 320 j to a low-temperature reservoir. if the efficiency of this engine is 56% of the efficiency of a carnot engine, what is the ratio of the low temperature to the high temperature in the carnot engine?
The ratio of the low temperature to high temperature of the Carnot engine is 2.38.
What is the efficiency of Carnot engine?The efficiency of the Carnot engine can be defined as the ratio of network done per cycle by the engine to the heat energy absorbed by the engine per cycle by the working substance from the source.
Efficiency = 1 - (Tlow/Thigh)
Heat absorbed by engine = 480J
Heat expelled by engine = 320J
Efficiency of the engine = 56% of efficiency of Carnot engine
The ratio of low temperature to high temperature in the Carnot engine.
Let's assume the efficiency of the Carnot engine is 'ηc' = 1 - T₂/T₁
Where, T₂ = Low temperature and T₁ = High temperature
To calculate the efficiency of the engine given, η = (Q1 - Q2)/Q1
η = (480 - 320)/480
η = 160/480
η = 1/3
η = 33.33%
Now, η = 56% × ηc
0.56ηc = 1/3ηc = (1/3)/0.56 = 0.58
As we already know, ηc = 1 - T₂/T₁
T₂/T₁ = 1 - ηc
T₂/T₁ = 1 - 0.58
T₂/T₁ = 0.42
T₁/T₂ = 1/0.42
T₁/T₂ = 2.38
Therefore, the ratio of low temperature to high temperature in the given Carnot engine with an efficiency of 56% will be about 2.38.
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