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2 years ago

You attach a cart with a mass of 0.92 kg to a spring and compress the cart-

Physics

1 answer:

Artyom0805 [142]2 years ago

8 0

Answer:

F = - K X force exerted by compressed spring

F X = 1/2 M V^2 energy (work) required to produce a speed V for mass M

K X * X = 1/2 M V^2 direction of F is the same as that of V

X^2 = M V^2 / K combining terms

X = (M / K)^1/2 V

X = (.92 / 130)^1/2 * 1.4 = .118 m

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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

A 2-Newton upward net force is being applied to a 10-kg object. What is the magnitude of the upward

Colt1911 [192]

Acceleration in m/s^2 = 2/10 = 0.2 m/s^2

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3 years ago

Electronegativity increases when atoms ___

notka56 [123]

your answer is.....

D. have a large atomic radius

although they also increase going from left to right so if D is incorrect, B might be your answer. it depends on context of the lesson.

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3 years ago

A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed

12345 [234]

7.17m/s glad I could help

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3 years ago

Accelerating car is a _______

Pavel [41]

Acceleration is the rate of change of velocity as a function of time. For example a car traveling at 50 km/hr starts to accelerate, 10 seconds after, its speed changes to 100 km/hr then the acceleration of the car during the time can be calculated as below: initial speed = 50 km/hr.

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2 years ago

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