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melisa1 [442]
3 years ago
15

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?
Physics
1 answer:
Katena32 [7]3 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 ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 2.0 s later. You ma
malfutka [58]

Answer:

4.9 m/s

Hope this helps! c:

Explanation:

3 0
3 years ago
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves
KIM [24]

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

4 0
3 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
3 years ago
Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?
sammy [17]

Question :

If a body acquires a charge of -0.02 C, has it gained or lost electrons? Many?

Solution :

We know, charge gained is shown by negative sign.

Since, charged acquired is given as -0.02 C .

Therefore, it is body has gained electrons.

Now, number of electrons is given by :

n = \dfrac{net\ charge}{charge \  on \ one \  electron}\\\\n = \dfrac{-0.02}{-1.60 \times 10^{-19}}\\\\n = 1.25\times 10^{17}

Hence, this is the required solution.

7 0
3 years ago
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc
sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

\overrightarrow{v}_{s,w}=6.5 \widehat{j}

velocity of water with respect to earth = 1.5 m/s at 40° north of east

\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}

\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j}  \right )

\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}

The magnitude of the velocity of ship relative to earth is \sqrt{1.15^{2}+5.54^{2}} = 5.66 m/s

5 0
3 years ago
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