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3 years ago

What is the displacement of the object from 12 seconds to 16 seconds

Physics

1 answer:

spayn [35]3 years ago

8 0

Answer:

<em>The displacement of the object is -8 m</em>

Explanation:

<u>Displacement</u>

The displacement of a moving object can be calculated as the area under (or above) the graph of velocity vs time.

If the area is below the y-axis, then the displacement is negative. Otherwise is positive.

It's important to differentiate displacement from distance. Displacement takes into consideration the direction of the movement. Distance does not and it's always positive.

From the graph provided, we can see the velocity from t=12 s from t=16 s is negative, and the displacement will also be negative.

The displacement is calculated as the area of the triangle with base b=16-12= 4 seconds and height = -4 m/s, thus:

$\displaystyle D = \frac{4*(-4)}{2}=-\frac{16}{2}=-8\ m$

The displacement of the object is -8 m

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If the average distance between bumps on a road is about 10 m and the natural frequency of the suspension system in the car is a

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When you hit a bump every 0.9 seconds.

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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.

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3 years ago

what will the stopping distance be a a 3000-kg car if -3000N of force are applied when the car is traveling 10 m/s

Inga [223]

Answer:

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<h3>What is a series?</h3>

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The formula that can be used to obtain a convergent series is given by;

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Learn more about a convergent series:brainly.com/question/15415793?

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2 years ago