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Paladinen [302]

3 years ago

5

In a microwave oven, electrons describe circular motion in a magnetic field within a special tube called amagnetron; as you'll l

earn in Chapter 29, the electrons' motion results in the production of micowaves. The electrons circle at a frequency of 2.15 GHz . The magnetron can accommodate electron orbits with maximum diameter 2.62 mm .
Part A

What's the magnetic field strength?

Part B

What's the electrons' energy in eV.

1 answer:

QveST [7]3 years ago

5 0

Answer:

The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

Explanation:

Given that,

Diameter = 2.62 mm

Frequency = 2.15 GHz

(A). We need to calculate the magnetic field strength

Using formula of the magnetic field strength

$B=\dfrac{2\pi mf}{e}$

Where, f = frequency

e = charge of electron

Put the value into the formula

$B=\dfrac{2\times3.14\times9.1\times10^{-31}\times2.15\times10^{9}}{1.6\times10^{-19}}$

$B=0.077\ T$

(B). We need to calculate the energy of electron

Using formula of energy

$E=\dfrac{1}{2}m(r\omega)^2$

$E=\dfrac{1}{2}\times9.1\times10^{-31}\times(1.31\times10^{-3}\times2\pi\times2.15\times10^{9})^2$

$E=1.4249\times10^{-16}\ J$

The energy in eV

$1 eV=1.6\times10^{-16}\ J$

$E=\dfrac{1.4249\times10^{-16}}{1.6\times10^{-16}}$

$E=0.8906\ eV$

Hence, The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

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A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i

Paha777 [63]

Answer:

A) $V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) $d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

$W_{d} = W_{a} - W_{w}$

Where:

$W_{a}$ : is the weight of the block in the air = 20.1 N

$W_{w}$ : is the weight of the block in the water = 15.3 N

$W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N$

Now, the mass of the water displaced is:

$m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg$

The volume of the block can be found using the mass of water displaced and the density of the water:

$V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) The density of the block can be found as follows:

$d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

I hope it helps you!

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3 years ago

In a “minute to win it” game, cards are placed between cups to stack them. The contestant then pulls the card out in hopes that

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Answer:

There is no friction between the card and the cup.

Explanation:

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3 years ago

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A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes

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Answer

D) burning a candle

Explanation

When burning a candle no new substance is form.

We have both physical and chemical change occuring.

Physical part: Melting of the solid wax and evaporation of the liquid forms the physical change.

Chemical part: burning of the wax vapour forms the chemical change.

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2 years ago

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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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