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olga nikolaevna [1]
3 years ago
8

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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If a force of 12 N is applied to an object
Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{12}{4}  \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

6 0
3 years ago
What is the density of this liquid that has a mass of 300g and volume 600ml.
8_murik_8 [283]

Answer:

0.5

Explanation:

D = M / V

You divide 300g by 600ml and you get 0.5

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8 0
3 years ago
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During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun
Ray Of Light [21]

Answer:

At the closest point

Explanation:

We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"

In this problem, we have a comet orbiting around the Sun:

- Its closest distance  from the Sun is 0.6 AU

- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: A\propto vr, therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster (A\propto vr, if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.

7 0
3 years ago
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

MV=mv

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

V = \frac{mv}{M}

Also we have that m/M = 1/7 times

On the other hand we have from law of conservation of energy that

W_f = KE

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

F_f*S = \frac{1}{2}MV^2

\mu M*g*S = \frac{1}{2}MV^2

\mu g*S = \frac{1}{2}( \frac{mv}{M})^2

\mu = \frac{1}{2} (\frac{1}{7}v)^2

\mu = \frac{1}{98}v^2

\mu = \frac{1}{g(98)(5.1)}v^2

Here we can apply the law of conservation of energy for light mass, then

\mu mgs = \frac{1}{2} mv^2

Replacing the value of \mu

\frac{1}{g(98)(5.1)}v^2  mgs = \frac{1}{2}mv^2

Deleting constants,

s= \frac{(98*5.1)}{2}

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7 0
3 years ago
You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe
NARA [144]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

I_{dB}=10\log(\dfrac{I}{I_{0}})

For one amplifier,

I_{1}=10\log(\dfrac{130}{I_{0}})...(I)

For other amplifier,

I_{2}=10\log(\dfrac{200}{I_{0}})...(II)

For difference in dB levels

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})

I_{2}-I_{1}=10\log(\dfrac{200}{130})

I_{2}-I_{1}=1.870\ dB

Hence, The sound level will be 1.870 dB louder.

7 0
3 years ago
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