Ask question

Ask question

All categories

3 months 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?

1 answer:

kicyunya [3.2K]3 months 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.

You might be interested in

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

5 0

3 months ago

A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

kicyunya [3294]

Answer:

$\Delta T = 0.81 ^oC$

Explanation:

According to the principle of energy conservation

all kinetic energy will change into thermal energy to increase its temperature

$\frac{1}{2}mv^2 = ms\Delta T$

Next, divide both sides by the object's mass

$\frac{1}{2}v^2 = s\Delta T$

the resulting temperature change is expressed as

$\Delta T = \frac{v^2}{2s}$

$\Delta T = \frac{25^2}{2\times 387}$

$\Delta T = 0.81^oC$

3 0

2 months ago

If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius

ValentinkaMS [3465]

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

8 0

3 months ago

Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be loca

Keith_Richards [3271]

Response:

1) An observer in B 'perceives the two events occurring at the same time

2) Observer B recognizes that the events happen at different times

3) Δt = Δt₀ /√ (1 + v²/c²)

Clarification:

This scenario illustrates the concept of simultaneity in special relativity. It is important to keep in mind that light's speed remains constant across all inertial frames

1) Since the events are stationary within the frame S ', they propagate at the constant speed of light, resulting in them reaching observation point B'—located equidistantly between both events—simultaneously

Thus, an observer in B 'observes the two events occurring at the same time

2) For an observer B situated within frame S attached to the Earth, both events at A and B appear to take place at the same moment. However, the event at A covers a shorter distance, while the event at B travels a longer distance, since frame S 'is in motion at velocity + v. Hence, with a constant speed, the event covering the lesser distance is perceived first.

Consequently, observer B perceives that the events do not occur simultaneously

3) Let's determine the timing for each event

Δt = Δt₀ /√ (1 + v²/c²)

where t₀ represents the time in the S' frame, which remains at rest for the events

8 0

2 months ago