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Shalnov [3]
3 years ago
11

The density of pure solid copper is 8.94 g/cm3. What volume does 5.0 kilograms of copper occupy? Leave your answer in the approp

riate number of significant figures and the unit for volume in cm3 and m3
Physics
1 answer:
azamat3 years ago
4 0

We have that the volume that 5.0 kilograms of copper occupy is

V=0.001788cm^3

From the question we are told

The density of pure solid copper is 8.94 g/cm3.

volume does 5.0 kilograms of copper occupy

Generally the equation for the Density  is mathematically given as

\rho=\frac{m}{V}\\\\Therefore\\\\V=\frac{m}{\rho}\\\\V=\frac{8.94}{5.0*1000}

V=0.001788cm^3

Therefore

the volume that 5.0 kilograms of copper occupy is

V=0.001788cm^3

In conclusion

The volume that 5.0 kilograms of copper occupy is

V=0.001788cm^3

For more information on this visit

brainly.com/question/13338592?referrer=searchResults

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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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As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. She then turns around and walks nor
Anastaziya [24]

Answer:

Option A

Solution:

As per the question:

The distance covered by the woman in the North direction, d = 3000 m

Time taken to travel in North direction, t = 25.0 min = 1500 s

Velocity of woman in the south direction, v = 2.00 m/s

Time taken in the south direction, t' = 60.0 min = 3600 s

Now,

The distance covered in the south direction, d' = vt' = 2.00\times 3600 = 7200\ m

Now, the total displacement is given by:

D = d' - d = 7200 - 3000 = 4200 m in South

(a) Average velocity of the woman in the whole journey is given by:

v_{avg} = \frac{Total\ displacement}{Total\ time} = \frac{4200}{t + t'}

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