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hichkok12
15 days ago
13

A magnetron in a microwave oven emits electromagnetic waves with a frequency f-2450 MHz. What magnetic field strength is require

d for electrons to move in circular paths with this frequency?
Physics
1 answer:
kicyunya [1K]15 days ago
6 0

Answer:

Magnetic field strength, B = 0.073 Tesla

Explanation:

The problem gives that a magnetron in a microwave oven emits electromagnetic waves at a frequency of 2450 MHz, f=2450\times 10^6Hz.

We need to determine the magnetic field strength that allows electrons to orbit with this frequency.

The applicable formula is:

B=\dfrac{m\omega}{q}

Here, q represents the electron charge and m its mass.

B=\dfrac{9.1\times 10^{-31}\times 2\pi (2450\times 10^6)}{1.6\times 10^{-19}}

Calculating, we find B = 0.073 Tesla.

Therefore, this is the required magnetic field.

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Response: Numerous elements can be found, all situated within the same vertical column as bromine.

Explanation:

Elements are organized by their atomic numbers on the periodic table. Those in the same vertical column (known as groups) exhibit the same valence electron configurations, resulting in similar chemical characteristics. Consequently, there are numerous elements sharing analogous chemical properties grouped with Bromine.

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a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,
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Consideremos la gravedad g como 9.8 m/s² y despreciemos la resistencia del aire.

La velocidad inicial vertical del guijarro es nula.
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Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnit
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Answer:

The kinetic energy is higher for the first cart.

Explanation:

For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to

(2kg)v = 10kg\: m/s

resulting in

v = 5m/s.

Consequently, the kinetic energy for the 3kg cart ends up as

K.E.  = \dfrac{1}{2}mv^2

= \dfrac{1}{2}(2kg)(5m/s)^2 =  25J

\boxed{K.E = 25J}

indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.

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15 days ago
A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
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The overall force acting on the vehicle is zero

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Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

  • The weight component acting perpendicular to the slope, mgcos \theta, directed into the slope
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Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

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The car moves at a constant speed in this direction, indicating that its acceleration is zero.

a=0

Thus, according to Newton's second law,

F=ma

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F=0

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In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric
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To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

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The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

g = \frac{GM}{R^2}

Where G represents the gravitational constant and M denotes the mass of the planet.

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now, let’s compute the time value.

h = \frac{1}{2} gt^2

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t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

Ultimately, the time for the rock to hit the surface is t = 87.58s.

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