Answer:
D = 2.38 m
Explanation:
This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit
a sin θ = m λ
Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle
θ = λ / a
Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant
θ = 1.22 λ / D
Where D is the circular tightness
Let's apply this equation to our case
D = 1.22 λ / θ
To calculate the angles let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (4.30 10⁻² / 140 10³)
θ = tan⁻¹ (3.07 10⁻⁷)
θ = 3.07 10⁻⁷ rad
Let's calculate
D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷
D = 2.38 m
Answer:
If she ___(be) rich, she _____ (buy) a new sports car.
Group of answer choices
is/buys
be/buy
were/would buy
were/bought
Answer:
B. - 0.328
Explanation
Potential Energy:<em> This is the energy of a body due to position.</em>
<em>The S.I unit of potential energy is Joules (J).</em>
<em>It can be expressed mathematically as</em>
<em>Ep = mgh........................... Equation 1</em>
<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>
<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>
Substituting these values into equation 1
Ep = 0.00274×12.2×9.8
Ep = 0.328 J.
Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J
The right option is B. - 0.328
<em />
Answer:(a) 2.40 (b) horizontal distance. (c) 0.630. (d) 6.50
Explanation:that's all is talking about a speed and distance and time right
Answer:
Explanation:
Given that,
Frequency of a radio antenna is 1 MHz
Power, P = 21 kW
We need to find the the waves intensity 25 km from the antenna
. The object emits intenisty evenly in all direction. It can be given by :
So, the wave intensity 25 km from the antenna is 
.
