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3 years ago

Hii please help i’ll give brainliest if you give a correct answer please please hurry it’s timed

Physics

1 answer:

Zina [86]3 years ago

3 0

Answer: The answer is A for sure

Explanation:

That is, there will be no acceleration. If you are sitting at rest in a chair and the upward push of the chair is equal to the downward pull of gravity, you will stay at rest in the chair. ... You now have an unbalanced force acting on you and therefore, according to Newton's First Law, your motion is going to change.

Plz give brainlest :)

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What is the value of acceleration that the car experiences

Anni [7]

Can u show the problem ?
Acceleration= the rate and change of an object also cars go fast so I think it would be positive acceleration
Hope I helped

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3 years ago

What does atomic spectrum mean?

lapo4ka [179]

Answer:

The range of characteristic frequencies of electromagnetic radiation that are readily absorbed and emitted by an atom. The atomic spectrum is an effect of the quantized orbits of electrons around the atom

Additional Facts:

Atomic spectra can also be analyzed to determine the composition of objects
The frequency depends on the difference in energy between the orbitals. Explaining this phenomenon was crucial to the development of quantum mechanics
Occurs due to the fact are quantized at specific levels determined by the atomic number

3 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

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

here, $\vec{P_\lambda}$ 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,

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

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

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

8 0

3 years ago

A Styrofoam ball has just been shot straight up. Air resistance is not negligible (free body diagram)

timofeeve [1]

In a free body diagram for an object projected upwards;

the acceleration due to gravity on the object is always directed downwards.

the velocity of the object is always in the direction of the object's motion.

An object projected upwards is subjected to influence of acceleration due to gravity.

As the object accelerates upwards, its velocity decreases until the object reaches maximum height where its velocity becomes zero and as the object descends its velocity increases, which eventually becomes maximum before the object hits the ground.

To construct a free body diagram for this motion, we consider the following;