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sammy [17]
3 years ago
11

A circular radar antenna on a coast guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 ghz. two small boats

are located 5.00 km away from the ship. how close together could the boats be and still be detected as two objects
Physics
1 answer:
solmaris [256]3 years ago
5 0
By definition,
q = 1.22y/D

Where,
q = min. angle
y = wavelength
D = Aperture diameter = diameter of the antenna

At distance "x" from the antenna,
L =xq = 1.22xy/D
Where, L = Min. distance

But, y =c/f = (3*10^8)/(16*10^9) = 0.01875 m

Substituting;
L = 1.22*5*10^3*0.01875/2.1 = 54.46 m
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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

\\

\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

\\

☯ <u>Now, Finding the force exerted </u>

\\

\dashrightarrow\:\: \sf{F = ma}

\\

\dashrightarrow\:\: \sf{F = 40 \times 5}

\\

\dashrightarrow\:\: \sf{F = 200 \ N}

\\

☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

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