Answer:
D
Explanation:
he describes as he writes them down
Answer:
You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>
Explanation:
By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.
This is given by :

Where
represents force applied on the object ,
represents mass of the object and
represents the acceleration.
In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.
A) 750 m
First of all, let's find the wavelength of the microwave. We have
is the frequency
is the speed of light
So the wavelength of the beam is

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find

and so, the diameter is

B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

So the area is

And since the power is

The average intensity is
