I need help with the questions when it says what is the topic and rest

What would happen if the distance between the earth and the moon decreased

makkiz [27]

The gravitational force between earth and moon will increase.
by F = GMm/r^2
F is inversely proportional to r^2
when r decrease, F will increase.

6 0

3 years ago

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How many protons, neutrons, and electrons does a neutral atom of this element have? (round atomic mass to nearest whole number)

adelina 88 [10]

Answer:

The element "AI" has:

Protons: 13 Neutrons: 14 Electrons: 13

Have a great day.

8 0

3 years ago

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The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.

Vadim26 [7]

Answer:

7.9 $\frac{gr}{cm^3}$

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7 $cm^3$ ).

Density is $\frac{mass}{volume}$ .

So the density of that piece of metal is $\frac{55.3g}{7cm^3}$

Which leaves us with a final density of 7.9 $\frac{gr}{cm^3}$

6 0

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.

8 0

4 years ago

Which is more useful for storing thermal energy, the block of lead or the water in the calorimeter?

leonid [27]

I believe it is the block of lead

5 0

3 years ago