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The sun emits a light with maximum wavelength 510 nm, while another star X emits a light with maximum wavelength of 350nm. What is the ratio surface temperature of the sun and the star X?
To calculate the ratio of the surface temperatures of the Sun and star X, we use Wien’s displacement law: λ_max × T = b where: - λ_max is the wavelength at maximum radiation, - T is the temperature of the object, - b is Wien's constant (2.9 × 10⁶ nm·K). The ratio of the temperatures is given by: (T_Read more
To calculate the ratio of the surface temperatures of the Sun and star X, we use Wien’s displacement law:
λ_max × T = b
where:
– λ_max is the wavelength at maximum radiation,
– T is the temperature of the object,
– b is Wien’s constant (2.9 × 10⁶ nm·K).
The ratio of the temperatures is given by:
(T_sun / T_star_X) = λ_max_star_X / λ_max_sun
Substitute the given values:
– λ_max_sun = 510 nm,
– λ_max_star_X = 350 nm.
(T_sun / T_star_X) = 350 / 510 = 0.686
Thus, the ratio of surface temperatures is approximately 0.68.
The correct answer is 0.68.
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Suppose the sun expands so that its radius becomes 100 times its present radius and its surface temperature becomes half of its present value. The total energy emitted by it then will increase by a factor of
We can apply the Stefan-Boltzmann law to compute the change in the total energy emitted: E = σ T⁴ A Here, - E is the total energy emitted - σ is the Stefan-Boltzmann constant - T is the surface temperature - A is the surface area of the object Surface area of sphere is given as: A = 4π r² If the radRead more
We can apply the Stefan-Boltzmann law to compute the change in the total energy emitted:
E = σ T⁴ A
Here,
– E is the total energy emitted
– σ is the Stefan-Boltzmann constant
– T is the surface temperature
– A is the surface area of the object
Surface area of sphere is given as:
A = 4π r²
If the radius of the sun increases by a factor of 100 and the temperature decreases by half, we need to calculate the change in energy.
Let the initial energy be E₁, and the final energy be E₂.
Initial energy: E₁ = σ T₁⁴ (4π r₁²)
Final energy: E₂ = σ (T₁/2)⁴ (4π (100r₁)²)
Simplifying the ratio of the final energy to the initial energy:
(E₂ / E₁) = (T₁/2)⁴ (100²) = (1/16) (10000) = 625
Thus, the total energy emitted increases by a factor of 625.
The correct answer is 625.
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A black body at a high temperature T K radiates energy at the rate of E Wm⁻². When the temperature falls to T/2 K, the radiated energy will be
According to the Stefan-Boltzmann law, the energy radiated by a black body is given by: E = σ T⁴ where: - E is the radiated energy, - σ is the Stefan-Boltzmann constant, - T is the temperature in Kelvin. Now, when the temperature is reduced to T/2, the new energy radiated will be: E' = σ (T/2)⁴ = (σRead more
According to the Stefan-Boltzmann law, the energy radiated by a black body is given by:
E = σ T⁴
where:
– E is the radiated energy,
– σ is the Stefan-Boltzmann constant,
– T is the temperature in Kelvin.
Now, when the temperature is reduced to T/2, the new energy radiated will be:
E’ = σ (T/2)⁴ = (σ T⁴) / 16
So the radiated energy decreases by a factor of 16, meaning the new radiated energy is E/16.
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A metal rod at a temperature of 150°C, radiates energy at a rate of 20 W. If its temperature is increased to 300°C, then it will radiate at the rate of
According to the Stefan-Boltzmann law, the radiated energy of a black body is proportional to the fourth power of its temperature: E ∝ T⁴ Let E₁ be the energy radiated at temperature T₁ = 150°C = 150 + 273 = 423 K, and let E₂ be the energy radiated at T₂ = 300°C = 300 + 273 = 573 K. We know that: E₁Read more
According to the Stefan-Boltzmann law, the radiated energy of a black body is proportional to the fourth power of its temperature:
E ∝ T⁴
Let E₁ be the energy radiated at temperature T₁ = 150°C = 150 + 273 = 423 K, and let E₂ be the energy radiated at T₂ = 300°C = 300 + 273 = 573 K.
We know that:
E₁ / E₂ = (T₁ / T₂)⁴
Substitute the values:
20 / E₂ = (423 / 573)⁴
Solving for E₂:
E₂ = 20 × (573 / 423)⁴ ≈ 37.2 W
The correct answer is 37.2 W.
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What tools can be used for weeding and what alternatives can be used if these are unavailable?
Weeding tools like hand cultivators are highly effective for removing weeds with minimal disruption to surrounding plants. If cultivators are unavailable, gardeners can use their hands or improvise with locally available materials, such as sticks, small spades, or sharp-edged objects. These alternatRead more
Weeding tools like hand cultivators are highly effective for removing weeds with minimal disruption to surrounding plants. If cultivators are unavailable, gardeners can use their hands or improvise with locally available materials, such as sticks, small spades, or sharp-edged objects. These alternatives ensure that weeds are removed without damaging plant roots. Consistent weeding, with or without tools, is crucial for maintaining soil health and preventing competition for resources among plants.
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