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noname [10]
4 years ago
15

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface t

emperature T and 3.0 times the diameter of the hotter star.
(a) What is the temperature of the hotter star in terms of T?
(b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?
Physics
1 answer:
Blizzard [7]4 years ago
7 0

Answer:

Explanation:

Let hotter star has surface area of A . The cooler star would have surface area 9 times that of hotter star ie 9A , because its radius is 3 times hot star. Let temperature of hot star be T ₁.

Total radiant energy is same for both the star

Using Stefan's formula of black body radiation,

For cold star E = 9A x σ T⁴

For hot star E = A x σ T₁⁴

A x σ T₁⁴ = 9A x σ T⁴

T₁⁴ = (√3)⁴T⁴

T₁ = √3T .

b )

Let the peak intensity wavelength be λ₁ and λ₂ for cold and hot star .

As per wein's law

for cold star , λ₁ T = b ( constant )

for hot star  λ₂ √3T = b

dividing

λ₁ T / λ₂ √3T = 1

λ₂  / λ₁ =  1 / √3

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A particle of mass 10 g and charge 80 mC moves through a uniform magnetic field,in a region where the free-fall acceleration is .
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Answer:

B=1.223\frac{T\cdot m}{s}\frac{1}{v}

Explanation:

According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:

F_m=F_g

Here F_m=qvBsin\theta is the magnetic force and F_g=mg is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so \theta=90^\circ. Replacing and solving for B:

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One star might appear brighter than another star because
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Answer:

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A car is traveling at a constant speed along the road ABCDE shown in the drawing. Section Ab and DE are straight. Rank the accel
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Answer:

AB = DE <CD <BC

Explanation:

This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.

Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.

Let's apply these reasoning to our problem.

AB Path  

this path is straight and as they indicate that the constant speed the acceleration is zero

DE path

This path is straight and since the velocity is constant the zero steps

BC path

This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression

         a_{c} = v² / r

where r is the radius of the curve and the direction of acceleration is towards the center of the curve

CD path

This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less

              a_{BC} > a_{CD}

In  summary lower accelerations are

 AB = DE <CD <BC

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