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2 years ago

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The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit in a magnetic field of

9.6 x 10^−2 T at a frequency of 2.7 GHz , and as they do so they emit 2.7 GHz electromagnetic waves.
a) If the maximum diameter of the electron orbit before the electron hits the wall of the tube is 2.5 cm, what is the maximum velocity the electrons may have?

1 answer:

Katena32 [7]2 years ago

8 0

Answer:

Explanation:

For equilibrium of rotating electron in outermost circular path

Centripetal force = force on electron

m v² / r = B q v

v = B q r / m , v is velocity of electron , B is magnetic field, q is charge on electron , r is radius of circular path.

v = 9.6 x 10⁻² x 1.6 x 10⁻¹⁹ x 1.25 x 10⁻² / 9.1 x 10⁻³¹

= 2.1 x 10⁸ m /s

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(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

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Answer:

a) $P = 370.993\,kPa$ , b) $F = 25.948\,kN$

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

$P = P_{atm} + P_{man}$

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 370.993\,kPa$

b) The force exerted by the water is:

$F = (P - P_{atm})\cdot A$

$F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}$

$F = 25.948\,kN$

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2 years ago

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Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

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2 years ago

If you lived on Saturn, which planets would exhibit retrograde motion like that observed for Mars from Earth? (Select all that a

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Answer:

a) Earth

b) Mercury

c) Neptune

Explanation:

All the planets move around the sun in eastward direction, but few planet have retrograde rotation i.e in westward direction. Retrograde motion is just an apparent change in the movement of planet which means it only seems as if the planet are rotating in opposite direction. Retrograde movement of planet like Saturn, Jupiter and mars is not real. Hence, if a person lives on Saturn, then following planets will exhibit retrograde motion

a) Earth

b) Mercury

c) Neptune

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2 years ago

Recall the previous question and the scenario with Zamir and Talia finding their way through a maze. Why is their displacement t

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Sample Response: Zamir and Talia’s total distances are different because they walked different paths in the maze. Zamir took a longer path. However, they had the same displacement because they both ended at the same position.

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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2 years ago