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zheka24 [161]
3 years ago
5

Which of the following types of stars is the coolest? In Graph A, the curve peaks at 800 nm, in the red section of the visible l

ight spectrum. In Graph B, the curve peaks at 550 nm, in the green section of the visible light spectrum. In Graph C, the curve peaks at 450 nm, in the blue section of the visible light spectrum. In Graph D, the curve peaks at 300 nm, in the violet section of the visible light spectrum.
Physics
2 answers:
ASHA 777 [7]3 years ago
8 0

im no scientist but im pretty sure cool stars are warm colors and hot stars are cool colors, so red would be more cooler than blue.

Anettt [7]3 years ago
8 0

Answer:

Coolest star is corresponding to that graph which have maximum wavelength. so correct answer will be

In Graph A, the curve peaks at 800 nm, in the red section of the visible light spectrum.

Explanation:

As we know by Wein's Displacement law that

\lambda = \frac{b}{T}

here it shows that

\lambda = wavelength corresponding to maximum intensity

T = temperature

so here if wavelength and temperature is inversely depends on each other so we can say if wavelength corresponding to maximum intensity is small then its temperature must be high.

In graph A wavelength is 800 nm

In graph B wavelength is 550 nm

In graph C wavelength is 450 nm

In graph D wavelength is 300 nm

so here coolest star must be corresponding to graph A

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Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

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part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

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It is the most frequent type of lever in the human body and as an example we can put the action of the brachial biceps in the flexion of the elbow, where the biceps is inserted in the forearm between the elbow that is behind and the resistance that would be displaced towards the hand by the weight of the load attached to the weight of the forearm.

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