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Virty [35]
2 years ago
10

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.

8 0
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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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.

5 0
2 years ago
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