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3 years ago

In an electronic transition an atom can not emit what?

1 answer:

7 0

█ Question █

In an electronic transition, an atom cannot emit what?

█ Answer █

When an electronic transition is occurring, an atom cannot emit ultra-violet light.

Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia

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Place a pencil or pen in front of you on the surface that you are working on. In one paragraph, using your own words, describe t

Ede4ka [16]

Answer:

READ THIS FIRST!!!

I am not going to right the parragraph (it says to use your own words, they could trace it back to me if you use my words, and you would get it trouble).

But I will tell you the forces acting on the pencil:

Gravity (pulls on the pencil)

Your hand (pushing it up canceling out gravity)

And the electromagnetic force (holding the thing together)

Hope it helped,

Have a good day

4 0

2 years ago

What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?

Marta_Voda [28]

Answer:

The answer is $\frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709$ . Let's learn why.

Explanation:

Newton's law of universal gravitation says;

$F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}$

Here G is a universal gravitational constant and is measured experimentally.

Sun's gravitational pull on mercury is:

$F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }$

Therefore $F_{Sun-Mercury} = Gm_{sun} 98,4366$

Sun's gravitational pull on Earth is:

$F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}$

Therefore $F_{Sun-Earth} =Gm_{sun} 265,33$

As a result;

$\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709$

4 0

3 years ago

You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) how much work do you do on the can of pa

MariettaO [177]

(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

$W=mg\Delta h$

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and $\Delta h=1.8 m$ is the variation of height. Substituting the numbers into the formula, we find

$W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J$

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

$W=Fd$

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.

(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

$W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J$