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3 years ago

A large truck and a compct car are both driving down the

Physics

1 answer:

koban [17]3 years ago

7 0

Answer:

Explanation:

The friction force acting on the body depends on the mass of the body, here the mas of truck is very large as compared to the car, so the friction force acting on the truck is large then the car. So, the amount of heat generated by the truck is more than the car and the truck overheated more.

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Which way does wind blow

prisoha [69]

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3 years ago

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The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The

Umnica [9.8K]

Answer:

$v_{rms} =196.59 m/s$

Given:

Temperature, T = 3.13 K

molar mass of molecular hydrogen, m = 2.02 g/mol = $2.02\times 10^{-3}kg/mol$

Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

$v_{rms} = \sqrt{\frac{3TR}{m}}$

where

R = rydberg's constant = 8.314 J/mol-K

Putting the values in the above formula:

$v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}$

$v_{rms} =196.59 m/s$

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3 years ago

¿Qué distancia recorrió un avión que viajaba a 750 km/h después de 2 h y media de vuelo?

arsen [322]

Answer:

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Explanation:

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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 creates an ionic bond?

natta225 [31]

Ionic bonding is the bonding between a positive metal with a negative nonmetal (metals are always positive while non metals are opposite). The meeting of a metal with a non metal creates an ionic bond.

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3 years ago

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