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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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Se ha quemado magnesio (reacción con el oxígeno) y se obtuvieron 12g de óxido de magnesio (II). ¿Cuánto magnesio se quemó? ¿Qué

Sedaia [141]

Answer:

The first answer is 7,24g Mg

The second answer is 3,36L O2

Explanation:

Atomic masses: Mg= 24,3 ; O=16

Solution: MMgO = 40,3 g/mol

2 Mg + O2 => 2 MgO

1) 12 g MgO x (2x24,3 g Mg / 2x 40,3 MgO) = 7,24 g Mg

At first calculate moles from Oxygen:

12 g MgO x ( 1 mol from Oxygen/ 2 x 40,3 g MgO ) = 0,15 moles from Oxygen

then ....

0,15 moles from Oxygen x (22,4 lts / 1 mol from Oxygen ) = 3,36 lts

22.4 liters is the constant that we use in this equation, it is a value that is repeated since we are in the presence of a gas with ideal behavior, so it is considered that the normal molar volume of ideal gaseous substances is always 22.4 liters, estimated with a temperature of 0º C and a pressure of 1 atmosphere. (constant temperature and pressure)

7 0

3 years ago

The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30%

nika2105 [10]

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

$f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

now it is given that tension is increased by 30%

so here we will have

$T' = T(1 + 0.30)$

$T' = 1.30T$

now new value of fundamental frequency is given as

$f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}$

now we have

$f_o' = \sqrt{1.3}f_o$

so here percentage change in the fundamental frequency is given as

$change = \frac{f_o' - f_o}{f_o} \times 100$

% change = 14.02%

3 0

3 years ago

every object that has mass attracts every other object with a gravitational force. it has been proven that the size of the gravi

erastova [34]

Answer;

- This statement about matter and its behavior is best classified as a Law.

-It is the Law of Universal Gravitation.

Explanation;

-The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation.This attractive force always points inward, from one point to the other.

-The Law applies to all objects with masses, big or small. Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric.

3 0

3 years ago

Read 2 more answers

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference, $\frac{dC}{dt}=12 cm/s$

Magnetic field,B=0.5 T

We have to find the induced emf at time t=9 s

We know that induced amf,E= $\frac{Bd(A)}{dt}$

Area of circular coil,A= $\pi r^2$

$E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}$

Circumference of circular coil,C= $2\pi r$

$165=2\pi r$

$r=\frac{165}{2\pi}$

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s$

Radius of coil at time t=9 s

$r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m$

$1 m=100 cm$

E= $-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V$

Magnitude of induced emf=0.005 V

4 0

3 years ago

Read 2 more answers

A typical tire for a compact car is 22 inches in diameter. If the car is traveling at a speed of 60 mi/hr, find the number of re

Softa [21]

Answer:

rpm= 916.7436 rev/min

Explanation:

First determine the perimeter of the wheel, to know the horizontal distance it travels in a revolution:

perimeter= π×diameter= π × 22 inches × 0.0254(m/inche)= 1.7555m

Time we divide the speed of the car, which is the distance traveled horizontally over time unit, by the perimeter of the wheel that is the horizontal distance traveled in a revolution, this dates us the revolutions over the time unit:

revolutions per time= velocity/perimeter

velocity= (60 mi/hr) × (1609.34m/mi) = 96560m/h

revolutions per time= (96560.6m/h) / (1.7555m)= 55004.614 rev/hr

rpm= (55004.614 rev/hr) × (hr/60min)= 916.7436 rev/min

3 0

4 years ago