What is the frequency of 20 mm microwaves? A) 100 MHz B) 400 MHz C) 15 GHz D) 73 GHz E) 98 GHz

Physics

1 answer:

Lera25 [3.4K]3 years ago

3 0

Answer:

Explanation:

Given wavelength = 20 mm = 20 x 10⁻³ m

The microwaves are electromagnetic waves so their velocity will be equal to

3 x 10⁸ m /s.

Now frequency of microwave = velocity / wavelength

= 3 x 10⁸ / 20 x 10⁻³ = 1.5 x 10¹⁰ = 15 x 10⁹ = 15 GHz.

You might be interested in

Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv

Anuta_ua [19.1K]

This is a good time to review Newton's 3rd law of motion:
"For every action, there is an equal and opposite reaction."

Gravitational force always acts in pairs.
Whatever force the Earth attracts something with,
the thing attracts the Earth with exactly the same force.

If Earth attracts a person with a gravitational force of 7.0 × 10² newtons,
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.

Your weight on Earth is the same as the Earth's weight on you !

5 0

3 years ago

What is Newton’s second law of motion

ddd [48]

"F = dP/dt. The net force acting on an object is equal to the rate at which its momentum changes."

These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is

"F = m·A. The net force on an object is equal to the product of the object's mass and its acceleration."

The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.

8 0

3 years ago

Read 2 more answers

Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as h

Arturiano [62]

Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.

So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle

As we know that arc length is given as

$L = R\theta$