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3 years ago

According to the spectrum seen below, which set of photons were absorbed by the element?

Physics

2 answers:

Effectus [21]3 years ago

5 0

The correct choice is ' D ' .

These are the numbers that are missing from the set of blocks at
the top, and also the bars missing from the graph in the middle.

I have no idea what the strip at the bottom labeled "full energy spectrum"
is trying to tell us.

Naddika [18.5K]3 years ago

5 0

Answer: Hello there!

you can see that the missing gray bars are at 2ev, 5ev, 8ev, and 15ev, these are the ranges where the photons were not detected (or absorbed by the element)

this is because in this kind of experimets you put the detector behind the element, then the photons that go through it are detected (and not absorbed) and the ones absorbed by the element never reach the detector, and are not detected.

this means that the right answer is D

The red green and blue are (almost) representing the visible spectrum of the energy of the photon, where red is between 1.6ev and 2ev, and violet is between 2.75ev end 3.26ev

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Katena32 [7]

v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s , hope this helps:)

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3 years ago

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How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at

tatuchka [14]

Answer:

0.66

Explanation:

By using the formula

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

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Nina [5.8K]

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3 years ago