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3 years ago

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A rectangular block of copper metal has a mass of 100 g. The dimensions of the block are 2 cm by 2 cm by 2 cm. From this data, w

hat is the density of copper?

2 answers:

lina2011 [118]3 years ago

6 0

Can I have some more detail

shusha [124]3 years ago

4 0

I think it is 4 copper

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In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell

Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

8 0

3 years ago

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baherus [9]

The answer is c to take a pictures with the camera in the metal hall

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3 years ago

A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how

Leona [35]

<span>345/115=3 3*24=72 the answer is B. 72 </span>

8 0

3 years ago

Read 2 more answers

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

4 years ago

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

$f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}$ .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

$f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}$ .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

7 0

3 years ago