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3 years ago

What does it mean when there is a curved line going upwards on a graph?science 8th grade :)

Physics

1 answer:

yulyashka [42]3 years ago

8 0

Answer: it is the asymptote

Explanation: a line that continually approaches a given curve but does not meet it at any finite distance.

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The sun, like all stars, releases energy through nuclear fusion. In this problem, you will find the total number of fusion react

sattari [20]

Answer:

Explanation:

DetaM=4 x 1.02875 - 4.002603

DetaM= 0.028697u

Using E= mc²

= 0.028697 x 1.49x*10^-10

= 4.2x10^-12J

4 0

3 years ago

Viewed moving objects from pond water under his microscope and named them “animalcules”.

galben [10]

Anton Von Leeuwenhoek, in the early 1600s, saw these tiny microbes and called them "animalcules" and "wee beasties".

5 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

Imagine a ringing bell set inside a sealed glass jar. Once all the air is removed and a vacuum is crated, the ringing sound is n

kenny6666 [7]

working...

Sound wave needs medium to travel

as energy which travels in this wave is because of transfer from one particle to another particle

If there is no medium then energy can not be transferred and sound wave will not travel

so in vacuum we can not listen sound

similarly here air is removed it means there is no medium inside the jar to travel the sound and hence we can not hear it

Option B is correct

Without air, the sound waves cannot travel to the ear.

5 0

3 years ago

Please help on this one?

bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0

2 years ago